Derivation and Regularization of Corrections to Propagators and Vertices for Cubically Interacting Scalar Field Theories

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DERIVATION AND REGULARIZATION OF CORRECTIONS TO PROPAGATORS AND VERTICES FOR CUBICALLY INTERACTING SCALAR FIELD THEORIES

A Thesis by

Richard Traverzo Bachelor of Science, Wichita State University, 2010

Submitted to the Department of Mathematics, Statistics, and Physics and the faculty of the Graduate School of Wichita State University in partial fulﬁllment of the requirements for the degree of Master of Science

The following faculty members have examined the ﬁnal copy of this thesis for form and content, and recommend that it be accepted in partial fulﬁllment of the requirement for the degree of Master of Science with a major in Mathematics.

Thomas DeLillo, Committee Chair

Holger Meyer, Committee Member

Thalia Jeﬀres, Committee Member

iii ABSTRACT

This thesis, a mainly expository work, attempts to give all the calculational details for some popular models in scalar quantum field theory that serve a mostly pedagogical purpose. Starting with a real scalar field with a given Lagrangian in classical field theory, standard quantization results given in most introductory QFT texts are derived with as many explicit calculations (which are often left to the reader) as possible. After presenting this development, we gather specific calculations that model scattering and decay processes. In many texts, the author(s) simply gives one or two calculations as an example. Here we organize several different calculations in a single work.

iv ACKNOWLEDGEMENTS

There are many people who I would like to thank for helping make this thesis possible.

First let me extend my gratitude to Dr. Tom DeLillo, my thesis advisor, for his direction and insights over the course of the past couple of years. The other members of my thesis committee, Dr. Thalia Jeﬀres and Dr. Holger Meyer, also deserve recognition for their participation in the process of my oral defense. Other faculty members who oﬀered friendly and useful advice were Dr. Abdulhamid Albaid, Dr. Elizabeth Behrman, and Dr. Jason Ferguson, and I am grateful for having recieved such advice. For personal support I extend my thanks to my family: my mother Wanda, uncle Leo, aunt Betty, grandmother Angelita, and grandfather Leonardo. Thank you to my girlfriend Ariana for her continued support throughout this process. Of course the greatest thanks goes to God who makes all things possible.

v PREFACE

The idea for this masters thesis originated out of Tom DeLillo’s 2009 Quantum Field

Theory course. He suggested that a good possible topic for the preliminary calculations and formalism for scalar quantum ﬁeld theory. Complete details for most theories are often scattered over several sources, therefore one objective of this thesis is to lay out all the calcu-

lational details of a popular model, or ”toy” theory, for students of physics or mathematics as preparation for more complicated physical theories (like Quantum Electrodynamics) including spin or charge. These details are also laid out as reference for a more rigorous

mathematical development of QFT; see eg. [1][2]. These calculations when combined into a manuscript could also be used as a guide for a student embarking on a course in QFT or even be used as a set of notes accompanying the standard texts of the ﬁeld.

With this in mind, the reader is assumed to be familiar with introductory quantum mechanics at the level of [3]. Also some basic special relativity is required. Though often presented in a tensor notation we forego this method in a general sense. Four-vectors, how-

µ ever, will be denoted in the typical sense: xµ = (x0, −x1, −x2, −x3) and x = (x0, x1, x2, x3) which are given by the metric signature (1, −1, −1, −1). In the introductory development

µ 2 2 2 2 we make use of the Einstein summation convention x xµ = x0 − x1 − x2 − x3.

vi TABLE OF CONTENTS

Chapter Page

1 INTRODUCTION...... 1

2 CLASSICAL FIELD THEORY...... 3

2.1 Lagrangian Mechanics; the Euler Lagrange Equations...... 3 2.2 The Euler-Lagrange Equations for Fields...... 5 2.3 Hamiltonian Mechanics...... 6 2.4 The Real Scalar Field...... 7 2.5 A Scalar Field Theory with Interactions...... 8

3 QUANTIZATION OF THE CLASSICAL KLEIN-GORDON FIELD...... 10

3.1 Obtaining the Hamiltonian and Momentum...... 13 3.2 Constructing the Fock Space and Evaluating the Energy and Momentum of its Elements...... 15 3.3 Obtaining a Time-Dependant Form of φ ...... 20 3.4 Quantizing the ABB and ABC-Theory...... 22 3.5 Calculating h0|ap, φ(x)|0i ...... 23 3.6 Derivation of the Feynman Propagator...... 25

4 PERTURBATION THEORY...... 29

4.1 Deriving the perturbation expansion...... 29 4.2 Expressing φ(x) in Terms of φI (x)...... 30

5 TREE-LEVEL AMPLITUDES IN ABC AND ABB THEORY...... 32

5.1 AB-Scattering in ABC Theory...... 32 5.1.1 The Zeroth Order Term in Expansion of g ...... 32 5.1.2 The First Order Term in Expansion of g ...... 33 5.1.3 The Second-Order Term in Expansion of g ...... 35 5.2 C → A + B Decay...... 43 5.3 A → B + B Decay...... 47

6 DIMENSIONAL REGULARIZATION...... 52

6.1 One-loop Correction to the A-Propagator in ABB Theory...... 52 6.2 One-loop Correction to the Vertex in ABC Theory...... 55 6.3 Fourth-Order Self Energy Diagram: Nested Divergence...... 56 6.4 Fourth-Order Self-Energy Diagram: Overlapping Divergence...... 59 6.4.1 Calculating the Boundary Contribution of the dη Integral...... 62 6.5 Calculating the integral contribution to the dη Integral...... 64

vii TABLE OF CONTENTS (continued)

Chapter Page

7 CONCLUSION AND FUTURE WORK...... 68

REFERENCES...... 71

viii LIST OF FIGURES

Figure Page

5.1 Zeroth-order term in the power series expansion of g...... 33

0 5.2 C-Exchange in AB-scattering. The qA refers to the pA in the discussion and

similarly for qB...... 37

0 5.3 AB Annihilation for ABC Theory. qA corresponds to pA and similarly for qB.. 41 5.4 A C-particle decaying into an A and a B-particle...... 43

5.5 An A-particle decaying into 2 B-particles...... 47 6.1 Self-Energy of The A-Particle in ABB Theory...... 53 6.2 One-loop correction to the vertex of ABC theory...... 55

6.3 Two-loop correction to the A-propagator: nested divergence...... 57 6.4 Two-loop correction to the propagator: overlapping divergence...... 59

ix CHAPTER 1

INTRODUCTION

Quantum ﬁeld theory is the attempt to combine quantum mechanics with special

relativity. In quantum mechanics, the (non-relativistic) Schrodinger equation which is second order in space and ﬁrst order in time gives rise to solutions which are interpreted as having positive energy. Unfortunately when we try to put forward this same interpretation in the

relativistic case (an equation both second order in space and time) we obtain “negative energy” solutions. To ameliorate this issue we extend relativistic quantum mechanics to relativistic quan-

tum field theory. φ is then interpreted not as a wavefunctions for a certain particle but as a field for infinitely many particles. Chapter 2 introduces field theory in the classical sense in preparation for quantization of the fields which takes place in Chapter 3. For this introductory material I follow mainly Peskin and Schroeder [4], although some material (especially the derivation of the Euler-Lagrange equation) is contained in Maggiore [5]. After the introductory development and a brief explanation of perturbation theory

(in Chapter 4), I present some basic interacting theories which are called the ABC and ABB-theories. The Feynman rules for ABC are stated in [6] and [7] and derived in [8]. The Feynman rules for the ABB theory are stated in [9], and to our knowledge have not been derived from a Lagrangian. We state this Lagrangian and derive the Feynman rule for a vertex. Chapter 5 is devoted to calculating scattering amplitudes, or S-matrix elements. In the laboratory any n number of particles collide with one another and turn into m (possibly diﬀerent) particles. The entries of the S-matrix hn|S|mi can be written in a perturbative expansion of some interaction parameter g and may be interpreted as the probability am-

plitude for the initial state to evolve into the particular ﬁnal state. These probabilities are

1 physically measured by decay rates Γ or scattering cross-sections σ both of which will not be considered here. This process of calculating amplitudes through evaluation of S-matrix elements works fine until higher-order contributions to the expansion are calculated. It turns out that almost all higher order calculations yield infinite results. We must then reinterpret what we mean by the parameters of the theory (mass, coupling, and field.) This is the subject of Chapter 6. Through a process known as dimensional regularization we are able to isolate the singular part of the integral and recover the finite part, discarding infinitesimal contributions. Covariant regularization is another method for isolating these infinite parts of the integrals (see [10]). It yields the same finite results (with different singular parts), but it will not be considered here.

2 CHAPTER 2

CLASSICAL FIELD THEORY

Before we are able to introduce quantum ﬁeld theory, we need to ﬁrst review the

basics of classical mechanics. [11] formulates mechanics in a number of useful ways including Poisson brackets and Hamilton-Jacobi theory (both of which are useful in the transition from classical to quantum mechanics.) Of most importance here, however, will be Lagrangian and

Hamiltonian mechanics. We will ﬁrst introduce the Lagrangian then deﬁne the Hamiltonian from it. In future chapters we will rely mainly on the Hamiltonian formalism.

2.1 Lagrangian Mechanics; the Euler Lagrange Equations

In classical mechanics, a system may be described by N generalized coordinates which themselves depend on time, a non-measurable quantity which we deﬁne as a real-valued, continuous parameter of the system. These are contructed as an attempt to capture the

measurable degrees of freedom of the system. The following derivations are following [5].

We denote these coordinates by qi = qi(t), i = 1, ..., N. How these coordinates interact with one another is captured by the Lagrangian. The Lagrangian L of this system is deﬁned by

L(q, q˙), a function of these generalized coordinates and their time derivatives which will be

dq typically denoted asq ˙ ≡ dt . As a system evolves from an initial configuration to a final one, we may define the action S of each possible path q(t) by

Z S = L(q, q˙)dt. (2.1) q(t)

The cornerstone of this classical theory is Hamilton’s Variational Principle, which states that the true path q(t) that the system takes is the one for which the action S is stationary. This is made mathematically formal by the methods of variational calculus and

3 setting the variation of S δS equal to 0.

Z tf 0 = δS = δ d4x L(q, q˙) dt tin Z tf = δL(q, q˙) dt tin Z tf h i = L(q + δq, q˙ + δq˙) − L(q, q˙) dt. tin

Now the variation of the Lagrangian is

X ∂L ∂L δL = δq + δq˙ , (2.2) ∂q i ∂q˙ i i i i

d so noting that δq˙i ≡ dt (δqi) we insert (2.2) into the formula above and integrate by parts to obtain

Z tf h X ∂L ∂L i 0 = δq + δq˙ + O((δq )2, (δq˙ )2) dt (2.3) ∂q i ∂q˙ i i i tin i i i tf Z tf h∂L d ∂Li ∂L X = − δq˙idt + δq(t) . (2.4) tin ∂qi dt ∂q˙i ∂q˙i i tin

The boundary vanishes since δq(tin) = δq(tf ) = 0, and we are left with

Z tf X h∂L d ∂Li 0 = − δq˙ dt. (2.5) ∂q dt ∂q˙ i tin i i i

Since the δqi(t)’s were arbitrary, we obtain at once the Euler-Lagrange equation ∂L d ∂L − = 0. (2.6) ∂qi dt ∂q˙i

As a motivation for (2.6) consider the mechanical energy Emech for a simple harmonic oscillator 1 1 E = mv2 + kx2. (2.7) mech 2 2 (2.7) “looks like” Hooke’s Law if one takes the derivative of PE with respect to x and the derivative of KE with respect to v (followed by a time derivative)

∂(PE) ∂(KE) d ∂(KE) = kx = mv ⇒ = ma. ∂x ∂v dt ∂v

4 The way to actually recover Hooke’s Law, F = −kx where k > 0 is the spring constant, is to deﬁne the Lagrangian L = T − V (where T = KE and V = PE) for the oscillator by

1 1 L = mx˙ 2 − kx2 (2.8) 2 2

for the generalized coordinate is q = x. Then we can apply (2.7) to obtain our result

∂L d ∂L 0 = − = −kx − mx.¨ (2.9) ∂qi dt ∂q˙i

2.2 The Euler-Lagrange Equations for Fields

We now make the transition from a system with a finite number of degrees of freedom to one with infinitely many. Consider a flexible string fixed at two endpoints on the x-axis.

Each point on the string can be thought of as being some distance away from the axis connecting the endpoints. Thus the string has infinitely many degrees of freedom. By thinking of the infinitely many degrees of freedom as a continuous variable, we create the concept of a field φ(x) where x = xµ = (x0, −x). Then for a system described by this field, the corresponding Lagrangian takes the form

Z 3 L = d xL(φ, ∂µφ) (2.10) where L is called the Lagrangian density. Since we are considering special relativity we will

∂ be using the four-vector notation throughout the text. In this case ∂µ = ( ∂t , −∇). Also, in keeping with custom L will be referred to simply as the Lagrangian. The action of a ﬁeld is then deﬁned by Z Z 4 S = dt L = d x L(φ, ∂µφ). (2.11)

This is an integral over all space-time. For the boundary conditions, instead of setting

δq(tin) = δq(tf ) = 0 like in the case of ﬁnitely many degrees of freedom, we assume that

µ φ, ∂µφ → 0 as x → ∞ suﬃciently fast so that the boundary terms may be neglected and the spacetime integrals are ﬁnite [5].

5 Once again the action S must be stationary for the actual (physical) ﬁeld φ Z 4 0 = δS = δ d xL(φ, ∂µφ) (2.12) Z 4 = d xδL(φ, ∂µφ) (2.13) Z 4 h∂L ∂L i = d x δφ + δ(∂µφ) . (2.14) ∂φ ∂(∂µφ)

We set δ(∂µφ) = ∂µ(δφ) and integrate by parts using a form of the divergence theorem for a finite volume of spacetime and taking the limit to all spacetime [5] to obtain Z Z 4 h∂L ∂L i ∂L 0 = lim d x + ∂µ δ(∂µφ) + dA δφ · nµ. (2.15) V →∞ V ∂φ ∂(∂µφ) ∂V ∂(∂µφ) Once again the boundary term goes vanishes as the volume approaches all of spacetime and we are left with Z 4 h ∂L ∂Li d x ∂µ − δφ = 0. (2.16) ∂(∂µφ) ∂φ Since the δφ is arbitrary, we finally obtain the Euler-Lagrange equations for fields

∂L ∂L ∂µ − = 0 . (2.17) ∂(∂µφ) ∂φ This is a 2nd-order PDE in our case.

2.3 Hamiltonian Mechanics

Next we pass to the Hamiltonian formalism. Note above that L was deﬁned in terms of q andq ˙: L = L(q, q˙). An alternative method to classical mechanics would be to deﬁne the

∂L conjugate momentum of a corresponding generalized coordinate by p ≡ ∂q˙ . The Hamiltonian for a given system is then deﬁned from its Lagrangian by H ≡ pq˙ − L and is a function of the generalized coordinates and their corresponding conjugate momenta H = H(p, q). The Euler-Lagrange equations are replaced by Hamilton’s equations

∂H ∂H −p˙i = q˙i = . (2.18) ∂qi ∂pi The same process is followed for ﬁelds by deﬁning the conjugate momentum by

∂L ∂ Z p(x) ≡ = L(φ(y), φ˙(y)d3y. (2.19) ∂φ˙ ∂φ˙(x)

6 Inverting the derivative with the integral we are able to obtain the deﬁnition for the conjugate

momentum density π(x) by ∂L π(x) ≡ . (2.20) ∂φ˙ The Hamiltonian H and corresponding Hamiltonian density H is then deﬁned by

Z Z H = d3x(π(x)q ˙(x) − L) = d3xH. (2.21)

It is worth noting that in this thesis we use the Lagrangian to deﬁne the Hamiltonian and

stay with the Hamiltonian formalism. It appears that the Hamiltonian is well suited for perturbation theory, whereas the Lagrangian formalism is best used in the path-integral formulation [5][4].

2.4 The Real Scalar Field

In this discussion we will only consider real scalar ﬁelds φ(x). From the Lagrangian

1 1 1 1 1 L = φ˙2 − (∇φ)2 − m2φ2 = (∂ φ)2 − m2φ2. (2.22) 2 2 2 2 µ 2

We can plug this into (2.17) to obtain the classical Klein-Gordon equation

∂2 − ∇2 + m2 φ = (∂ ∂µ + m2)φ = 0. (2.23) ∂t2 µ

µ 2 2 Note the summation notation on the right-hand side. For a general 4-vector xµx = t − x .

µ ∂2 2 In this case ∂µ∂ = ∂t2 − ∇ . From the Lagrangian we can determine the Hamiltonian. Noting that

∂L ∂ 1 1 1 π(x) = = φ˙2 − (∇φ)2 + m2φ2 = φ˙(x), (2.24) ∂φ˙(x) ∂φ˙(x) 2 2 2

we use the deﬁnition of H to obtain

H = πφ˙ − L (2.25) 1 1 1 = π2 − φ˙2 + (∇φ)2 + m2φ2 (2.26) 2 2 2 1 1 = π2 + (∇φ)2 + m2φ2 (2.27) 2 2

7 with the corresponding total Hamiltonian being

Z Z 1 1 1 H = Hd3x = π2 + (∇φ)2) + m2φ2 d3x. (2.28) 2 2 2

Each of the terms in H has a speciﬁc interpretation ([4], p. 17) as the energy cost for:

1 π2 ⇒ moving in time 2 (∇φ)2) ⇒ shearing in space 1 m2φ2 ⇒ existence of the ﬁeld. 2

2.5 A Scalar Field Theory with Interactions

Consider a physical theory describing 3 scalar particles as described in [6] and [8]. It is referred to as the ABC-theory. We will call them A, B, and C particles and the corresponding

ﬁelds will be denoted φA, φB, and φC , respectively. If they were non-interacting particles, the Lagrangian of the system would be written as

1 1 1 1 1 1 L = (∂ φ )2 − m2 φ2 + (∂ φ )2 − m2 φ2 + (∂ φ )2 − m2 φ2 (2.29) ABC 2 µ A 2 A A 2 µ B 2 B B 2 µ C 2 C C which contains 3 identical copies of (2.22). And if we apply the Euler-Lagrange equations to this we obtain 3 identical KG-equations.

µ 2 (∂µ∂ + mA)φA = 0 (2.30)

µ 2 (∂µ∂ + mB)φB = 0 (2.31)

µ 2 (∂µ∂ + mC )φC = 0. (2.32)

Now suppose we introduce an interaction between the particles by adding a term to

(2.29) which is a product of the three ﬁelds at a point in spacetime Lint = −gφA(x)φB(x)φC (x). With this simple interaction term the total Lagrangian becomes

1 1 1 1 1 1 L = (∂ φ )2 − m2 φ2 + (∂ φ )2 − m2 φ2 + (∂ φ )2 − m2 φ2 −gφ φ φ . (2.33) ABC 2 µ A 2 A A 2 µ B 2 A B 2 µ A 2 A A A B C

8 An application of the Euler-Lagrange equations gives

µ 2 (∂µ∂ + mA)φA = −gφBφC (2.34)

µ 2 (∂µ∂ + mB)φB = −gφAφC (2.35)

µ 2 (∂µ∂ + mC )φC = −gφAφB. (2.36)

We will also consider another physical theory: one describing only 2 particlse, an A and a B. We will call it the ABB-Theory, and it is discussed in [9]. The full Lagrangian is a bit simpler than the ABC

1 1 1 1 1 L = (∂ φ )2 − m2 φ2 + (∂ φ )2 − m2 φ2 − gφ φ2 . (2.37) ABB 2 µ A 2 A A 2 µ B 2 B B 2 A B

The Euler-Lagrange Equations for this Lagrangian give 2 equations of motion with diﬀerent interacting terms

µ 2 2 (∂µ∂ + mA)φA = −gφB (2.38)

µ 2 (∂µ∂ + mB)φB = −gφAφB. (2.39)

The equations of motion for ABC-Theory and ABB-Theory are systems of PDE’s, and a mathematical analysis to study existence and uniqueness of solutions could be undertaken. We will not pursue this here. Instead, we will quantize the free Klein-Gordon equation to

R 4 obtain non-interacting states and rely on a perturbative expansion of ei d xHint to compute probabilites of one free state evolving to another through the interaction.

9 CHAPTER 3

QUANTIZATION OF THE CLASSICAL KLEIN-GORDON FIELD

Now that we have the classical ﬁeld equation, we quantize the ﬁeld by promoting

φ(x) and π(x) to operators (acting on a Hilbert space) and imposing commutation relations between them. Recall that the commutator between two operators is deﬁned by

[A, B] := AB − BA. (3.1)

The commutation relations which will be imposed on φ and π are

[φ(x, t), π(y, t)] = iδ3(x − y) (3.2)

[φ(x, t), φ(y, t)] = [π(x, t), π(y, t)] = 0. (3.3)

Note that the commutation relations above are for the ﬁeld operators in the Heisenberg picture, where they depend on time. In fact, these are equal-time commutation relations.

Were we in the Schrodinger picture the fields, and hence the commutator, would be fixed for all time, since only the states depend on time in the Schrodinger picture. To “solve” this quantum field theory, we first must find an expression for the Klein-

Gordon equation in momentum space. We do this by ﬁrst expanding φ(x) itself as an integral over momentum space Z d3p φ(x, t) = eip·xφ(p, t). (3.4) (2π)3 Now the Klein-Gordon equation in position space may be written

∂ − ∇2 + m2 φ(x) = 0, (3.5) ∂t2

10 so we can replace the ﬁeld φ(x) in (3.5) with its momentum space expression (3.4) and write the KG-equation itself in momentum space by the following calculation

∂ 0 = − ∇2 + m2 φ(x) ∂t2 ∂ Z d3p = − ∇2 + m2 eip·xφ(p, t) ∂t2 (2π)3 Z d3p ∂ = − ∇2 + m2 eip·xφ(p, t). (2π)3 ∂t2

And since ∇2eip·x = (ip · ip)eip·x = −|p|2eip·x we can plug this into our expression to obtain

Z d3p ∂ 0 = + |p|2 + m2 eip·xφ(p, t). (2π)3 ∂t2

This of course yields our result, the Klein-Gordon equation in momentum space

∂ + |p|2 + m2 φ(p, t) = 0. (3.6) ∂t2

Since φ is real we must have solutions of the form

iωpt † −iωpt φ(p, t) ∝ ape or φ(p, t) ∝ ape

where the frequency ωp is given by

p 2 2 ωp = |p| + m . (3.7)

This takes the same form as the quantum harmonic oscillator (see [3] for a discussion of the quantum harmonic oscillator.) In this way we may think of each mode of the Klein-

† Gordon ﬁeld as a harmonic oscillator with its own raising and lowering operators (ap and

ap, repectively). For the ﬁeld φ(x) and its corresponding conjugate momentum density π(x) we have the following mode expansions

Z d3p 1 φ(x) = a eip·x + a† e−ip·x (3.8) 3 p p p (2π) 2ωp Z d3p rω π(x) = (−i) p a eip·x − a† e−ip·x , (3.9) (2π)3 2 p p

11 and if we allow p to take on negative values we obtain a more useful form for φ and π Z d3p 1 φ(x) = a + a† eip·x (3.10) 3 p p −p (2π) 2ωp Z d3p rω π(x) = (−i) p a − a† eip·x. (3.11) (2π)3 2 p −p The initial commutation relations for φ and π are actually equivalent to the commu-

† tation relation between ap and ap

† 3 3 [ap, aq] = (2π) δ (p − q) (3.12)

as is shown by the following calculation:

[φ(x), π(y)] = φ(x)π(y) − π(y)φ(x) Z d3p 1 Z d3q rω = (a + a† )ei(p·x) (−i) p (a − a† )ei(q·y) 3 p p −p 3 q −q (2π) 2ωp (2π) 2 Z d3q rω Z d3p 1 − (−i) p (a − a† )ei(q·y) (a + a† )ei(p·x) 3 q −q 3 p p −p (2π) 2 (2π) 2ωp Z 3 3 r d pd q −i ωq h † † i i(p·x+q·y) = 6 (ap + a−p)(aq − a−q) e (2π) 2 ωp Z 3 3 r d pd q −i ωq h † † i i(p·x+q·y) − 6 (aq − a−q)(ap + a−p) e (2π) 2 ωp Z 3 3 r d pd q −i ωq h † † † † i i(p·x+q·y) = 6 (ap + a−p)(aq − a−q) − (aq − a−q)(ap + a−p) e . (2π) 2 ωp The product of creation an annihilation operators in the integrand may be simpliﬁed to a sum of commutators:

† † † † † † † † (ap + a−p)(aq − a−q) − (aq − a−q)(ap + a−p) = [ap, aq] + [a−q, ap] + [a−p, aq] + [a−q, a−p]

= 0 − (2π)3δ3(p + q) − (2π)3δ3(q + p) + 0

= −2(2π)3δ3(p + q).

Inserting this result into the above expression we obtain the desired result Z 3 3 r d pd q −i ωq h 3 3 i i(p·x+q·y) [φ(x), π(y)] = 6 − 2(2π) δ (p + q) e (3.13) (2π) 2 ωp Z d3p = (i)eip·(x−y) (3.14) (2π)3 = iδ3(x − y). (3.15)

12 3.1 Obtaining the Hamiltonian and Momentum

Next we calculate the Hamiltonian of the system. Recall that

Z Z 1 1 1 H = H d3x = π2 + (∇φ)2 + m2φ2 d3x, (3.16) 2 2 2

so to express H in terms of a and a† we ﬁrst calculate the individual terms of H. For the

ﬁrst term we have

Z d3p rω Z d3q rω π2 = (−i) p a − a† eip·x (−i) q a − a† eiq·x (3.17) (2π)3 2 p −p (2π)3 2 q −q 3 3 √ Z d pd q − ωpωq = (a − a† )(a − a† )ei(p+q)·x. (3.18) (2π)6 2 p −p q −q

For the second term we have

Z d3p 1 ∇φ = ∇ a + a† eip·x (3.19) 3 p p −p (2π) 2ωp Z d3p 1 h i = a + a† ∇eip·x (3.20) 3 p p −p (2π) 2ωp Z d3p 1 = a + a† (ip)eip·x (3.21) 3 p p −p (2π) 2ωp

which, when squared, becomes

Z d3p 1 Z d3q 1 (∇φ)2 = a + a† (ip)eip·x a + a† (iq)eiq·x (3.22) 3 p p −p 3 p q −q (2π) 2ωp (2π) 2ωq Z d3pd3q 1 = (a + a† )(a + a† )(−p · q)ei(p+q)·x. (3.23) 6 p p p q q (2π) 2ωpωq The last term takes the same form as (∇φ)2 but with no −(p · q)

Z d3pd3q 1 φ2 = (a + a† )(a + a† )ei(p+q)·x. (3.24) 6 p p p q q (2π) 2ωpωq Now that we have the form for each of the terms of H in terms of creation and annihilation operators we can ﬁnally integrate to obtain a form for H (cf. [4], eq. 2.31).

Z Z d3pd3q H = d3x ei(p+q)·x (2π)6 √ 2 − ωpωq † † −p · q + m † † × (ap − a−p)(aq − a−q) + √ (ap + ap)(aq + aq) . 4 4 ωpωq

13 We ﬁrst evaluate the d3x integral to obtain a δ-function δ3(p + q) which will enforce q = −p

upon integration over q.

Z 3 √ 2 d p h− ωpω−p † † −p · (−p) + m † † i H = 3 (ap − a−p)(a−p − ap) + √ (ap + ap)(a−p + ap) (2π) 4 4 ωpω−p (3.25) Z d3p ω = p − a a + a a† + a† a − a† a† + a a + a a† + a† a + a† a† (2π)3 4 p −p p p −p −p −p p p −p p p −p −p −p p (3.26) Z d3p ω = p 2a a† + 2a† a (3.27) (2π)3 4 p p −p −p Z d3p ω = p 2a a† + 2a† a + 2a† a − 2a† a (3.28) (2π)3 4 p p −p −p p p p p Z d3p ω = p 4a† a + 2[a , a† ] (3.29) (2π)3 4 p p p p Z d3p 1 = ω a† a + [a , a† ] . (3.30) (2π)3 p p p 2 p p

† The [ap, ap] term is the zero-point energy, a c-number which is immeasurable. It can be eliminated if we redeﬁne the Hamiltonian with normal ordering. For any product of creation

† † and annihilation operators ap and ap the normal ordering operation takes all the ap’s and

puts them on the left side of the product leaving all the ap’s on the right side (standard commutation rules apply.) The operation is denoted with a pair of colons on either side of

† † the product. As an example : apap : = apap. Hence if we deﬁne our Hamiltonian density by

1 1 1 H = : π2 + (∇φ)2 + m2φ2 : . (3.31) 2 2 2

then we obtain the result we need

Z d3p H = ω a† a . (3.32) (2π)3 p p p

Next we follow a similar process to express the (total) momentum operator in terms

of creation and annihilation operators (cf. [4], eq. 2.33). From the momentum-energy tensor ([5], sec. 3.2.1) the total momentum operator is deﬁned as Z P = − d3xπ(x)∇φ(x). (3.33)

14 Substituting our expressions for π and ∇φ we obtain

Z Z d3p rω Z d3q 1 P = − d3x (−i) p a − a† eip·x a + a† (ip)eiq·x 3 p −p 3 p q −q (2π) 2 (2π) 2ωq Z 3 3 r Z d pd q 1 ωp † † 3 i(p+q)·x = − 6 (ap − a−p)(aq + a−q)q d xe (2π) 2 ωq Z 3 3 r d pd q 1 ωp † † 3 3 = − 6 (ap − a−p)(aq + a−q)q(2π) δ (p + q), (2π) 2 ωq and upon integration over q the δ-function enforces q = −p so we are left with

Z 3 r d p 1 ωp † † P = − 3 (ap − a−p)(a−p + ap)(−p) (2π) 2 ω−p Z d3p 1 = p(a a + a a† − a† a − a† a† ). (2π)3 2 p −p p p −p −p −p p

The ﬁrst and last terms vanish since they are odd in p, and if we redeﬁne P by a normal ordering of the integrand as was the case with H we are left with our result (cf. [4], eq. 2.33)

Z d3p P = p a† a . (3.34) (2π)3 p p

3.2 Constructing the Fock Space and Evaluating the Energy and Momentum

of its Elements

Now that we have our Hamiltonian H and momentum P operators expressed in terms

† of ap and ap we proceed to construct the mathematical space on which these operators will act. The resulting elements of this space will be interpreted as free, non-interacting multi- particle states.

Deﬁnition 1. The vacuum state, denoted by |0i, is the state that is destroyed when acted upon by the annihilation operator. Formally, for all p we have

ap|0i = 0. (3.35)

This state is normalized by h0|0i = 1. Upon hermitian conjugation of (3.35) we

† clearly also have h0|ap = 0. The elements of the space which will carry physical signiﬁcance are those obtained when the creation operator acts on |0i.

15 Deﬁnition 2. A one-particle state |pi is deﬁned by

p † |pi = 2Epap|0i. (3.36)

We may interpret |pi as a single relativistic particle propagating through space with momentum p. We can deﬁne an inner product on the space by

1/2 1/2 † hp|qi = h0|ap(2Ep) (2Eq) aq|0i

1/2 1/2 † † = (2Ep) (2Eq) h0|apaq − aqap|0i

1/2 1/2 † = (2Ep) (2Eq) h0|[ap, aq]|0i

1/2 1/2 3 3 = (2Ep) (2Eq) h0|(2π) δ (p − q)|0i

3 3 = 2Ep(2π) δ (p − q).

From this we can deﬁne the notion of a multi-particle state by ([5], eq. 4.9)

1/2 1/2 1/2 † † † |p1, p2, ..., pni ≡ (2Ep1 ) (2Ep2 ) ··· (2Epn ) ap1 ap2 ··· apn |0i. (3.37)

Now that we have the elements of our space constructed we proceed to operate on them with our energy and momentum operators, H and P. First note that, by deﬁnition of the vacuum state, H|0i = 0 and P|0i = 0

Z d3p H|0i = E a† a |0i = 0 (3.38) (2π)3 p p p Z d3p P|0i = pa† a |0i = 0. (3.39) (2π)3 p p

For a one-particle state we compute the energy and momentum as follows. First from H and

16 (3.36) we have

= Ep|pi.

Similarly for momentum we have

= p|pi.

We can also compute the energy and momentum for a two-particle state |p1, p2i.

Z d3q h H|p , p i = E a† a (2E )1/2(2E )1/2a† a† |0i 1 2 (2π)3 q q q 1 2 p1 p2 Z d3q = (2E )1/2(2E )1/2 E a† [a , a† a† ]|0i. 1 2 (2π)3 q q q p1 p1

Note that [A, BC] = [A, B]C + B[A, C] so that the commutator above reduces to

Z d3q (2E )1/2(2E )1/2 E a† [a , a† ]a† + a† [a , a† ] |0i. (3.40) 1 2 (2π)3 q q q p1 p2 p1 q p2

Once we have integrated out the δ-functions we are left with

1/2 1/2 † † † † (2E1) (2E2) E1ap1 ap2 + E2ap2 ap1 |0i. (3.41)

17 For our result we obtain H|p1, p2i = (E1 +E2)|p1, p2i. As expected, the Hamiltonian acting on a two particle state gives us the sum of the energies of the particles. A similar result

holds for P. To compute the energy or momentum for a more general state we ﬁrst need the following lemma for the commutator of a product of operators. The basic idea is that the

commutator “works its way through” the product with diﬀerent terms in a sum. For example,

[A, B1B2B3B4] simpliﬁes to

[A, B1B2B3B4] = [A, B1]B2B3B4 + B1[A, B2]B3B4 + B1B2[A, B3]B4 + B1B2B3[A, B4].

Lemma 1. Let A, B1,B2, ..., Bn be arbitrary operators. Deﬁne B0 = Bn+1 = 1. Then for n ≥ 2 n j−1 n+1 X Y Y [A, B1B2 ··· Bn] = Bi[A, Bj] Bk . (3.42) j=1 i=0 k=j+1 Proof. The proof is by induction. For n = 2 we have

[A, B1B2] = AB1B2 −B1B2A = AB1B2 −B1AB2 +B1AB2 −B1B2A = [A, B1]B2 +B1[A, B2]. (3.43)

If we assume by inductive hypothesis that (3.42) holds for [A, B1B2 ··· Bn], then we can compute the commutator for n + 1

n h Y i [A, B1B2 ··· BnBn+1] = A, Bi Bn+1 i=1 n n Y Y = A Bi Bn+1 − Bi Bn+1A i=1 i=1 n n n n Y Y Y Y = A Bi Bn+1 − Bi ABn+1 + Bi ABn+1 − Bi Bn+1A i=1 i=1 i=1 i=1 n n h Y i Y = A, Bi Bn+1 + Bi[A, Bn+1] i=1 i=1 n j−1 n+1 n X Y Y Y = Bi[A, Bj] Bk Bn+1 + Bl[A, Bn+1]Bn+2 j=1 i=0 k=j+1 l=0 n+1 j−1 n+2 X Y Y = Bi[A, Bj] Bk . j=1 i=0 k=j+1

18 Thus the result holds for the case of n + 1 and our proof is complete.

Using this result we can compute the energy and momentum of a general multi- particle state:

Proposition 1.

H|p1, p2,..., pni = (Ep1 + Ep2 + ··· + Epn )|p1, p2,..., pni. (3.44)

Proof.

n Z d3p Y H|p , p ,..., p i = E a† a (2E )1/2a† |0i 1 2 n (2π)3 p p p pi pi i=1 n n n Y Z d3p Y Y = (2E )1/2 E a† a a† − a† a |0i pi (2π)3 p p p pj pj p i=1 j=1 j=1 n n Y Z d3p h Y i = (2E )1/2 E a† a , a† |0i. pi (2π)3 p p p pj i=1 j=1 After applying the result of (3.42) we reduce the commutator to a sum of individual commutators of the form [a , a† ]. From this we evaluate these commutators as a series of p pi δ-functions and proceed to integrate them:

n n Y X Z d3p Y H|p , p ,..., p i = (2E )1/2 E a† (2π)3δ3(p − p ) a† |0i 1 2 n pi (2π)3 p p j pi i=1 j=1 i6=j n n n Y X Y = (2E )1/2 E a† |0i pi pj pi i=1 j=1 i=1 n n X Y = E (2E )1/2a† |0i pj pi pi j=1 i=1

= (Ep1 + Ep2 + ··· + Epn )|p1, p2,..., pni, where in the second line we have used the fact that all creation operators commute with one another. Thus we have arrived at our result.

A similar result which we state without proof holds for the spatial momentum of a multi-particle state:

P|p1, p2,..., pni = (p1 + p2 + ··· + pn)|p1, p2,..., pni. (3.45)

19 We have started from the classical Klein-Gordon equation as derived from the invari-

ant action and imposed canonical commutation relations between φ(x, t) and π(y, t) at equal times. Switching to a momentum space representation, the classical KG-ﬁeld was quantized by interpreting each mode of the ﬁeld as a quantum harmonic oscillator with the associated

† creation and annihilation operators (ap and ap.) We then expressed the Hamiltonian H

† and momentum P operators in terms of ap and ap. After constructing the Fock space by defining an element (with appropriate relativistic normalization) to be a product of creation operators acting on the initial “vacuum” element |0i, we operated on these states with H and P so that the appropriate interpretation of an arbitrary element of the Fock space as a multiparticle state with a definite energy and momentum may be applied. In this way we say that the (free) quantum field theory has been solved.

3.3 Obtaining a Time-Dependant Form of φ

An important physical quantity that will arise is the amplitude of a particle to prop- agate from spacetime points x to y. This process will be represented by

h0|φ(x)φ(y)|0i. (3.46)

This is a vacuum expectation value or vev of a product of two fields. These vev’s are in effect what are actually measured in the laboratory. Mathematically , it can be shown by The Reconstruction Theorem [2] that the field can be recovered (or reconstructed) from the

vacuum expectation values. In order to calculate this, the ﬁrst task is to convert our KG-ﬁeld

Z d3x 1 φ(x) = a eip·x + a† e−ip·x (3.47) 3 p p p (2π) 2Ep into a form which contains x and p, not just the spatial parts x and p (four-vectors x containing temporal and spatial components are denoted in the usual way as x = xµ =

(x0, −x1, −x2, −x3).) The following discusssion fills in the calculational details of the last paragraph of p. 25 in Section 2.4 of [4]. To compute the time-dependance of φ(x) we switch to the Heisenberg picture by defining the time-dependant field operator φ(x) in terms of the

20 time-independant φ(x)

φ(x) ≡ φ(t, x) = eiHtφ(x)e−iHt (3.48) Z d3p 1 = eiHta e−iHteip·x + eiHta† e−iHte−ip·x . (3.49) 3 p p p (2π) 2Ep

Since promoting φ(x) to an operator essentially meant promoting ap and apd to operators, the problem of determining the time-dependance of φ(x) ultimately rests on the time-

† dependance of ap and ap. To determine this we ﬁrst evaluate the commutator [H, ap] by the following calculation: h Z d3q i [H, a ] = E a† a , a p (2π)3 q q q p Z d3q = E [a† a , a ] (2π)3 q q q p Z d3q = − E (2π)3δ3(p − q)a (2π)3 q q

= −Epap.

† † By a similar calculation [H, ap] = Epap so that we have the following two identities

† † Hap = ap(H − Ep) Hap = ap(H + Ep). (3.50)

2 2 To ﬁnd the quantity H ap we compute [H , ap] using the commutator [H, ap] and obtain

2 2 2 [H , ap] = H ap − apH

2 2 = H ap − HapH + HapH − apH

= H[H, ap] + [H, ap]H

2 from this we can deduce the expression for H ap by letting

2 [H , ap] = H[H, ap] + [H, ap]H

2 2 H ap = apH − H(Epap) − (Epap)H

2 = apH − Epap(H − Ep) − (Epap)H

2 2 = apH − ap(2EpH) + apEp

2 = ap(H − Ep) .

21 † 2 † † 2 Similarly for ap we have H ap = ap(H + Ep) . Clearly the general terms take the form

n n n † † n H ap = ap(H − Ep) H ap = ap(H + Ep) . (3.51)

iHt −iHt Using this general form we can then compute e ape by expanding out the exponential on the left-hand side and applying (3.51);

1 eiHta e−iHt = 1 + (iHt) + (iHt)2 + ··· a e−iHt p 2 p 1 = a + iHa t + (−i)2H2a t2 + ··· e−iHt p p 2 p 1 = a + ia (H − E ) + (−i)2a (H − E )2t2 + ··· e−iHt p p p 2 p p

i(H−Ep)t −iHt = ape e

−iEpt = ape .

† And with a similar calculation for ap we obtain the time evolution of our creation and

† annihilation operators ap and ap ([4], eq. 2.46)

iHt −iHt −iEpt iHt † −iHt † iEpt e ape = ape e ape = ape (3.52)

Plugging in our results of (3.52) into φ(x) we obtain its explicit form

Z d3p 1 φ(x) = a e−ip·x + a† eip·x . (3.53) 3 p p p (2π) 2Ep Note also that the time evolution of the conjugate momentum ﬁeld is deﬁned in a similar manner by π(x) = eiHtπ(x)e−iHt so that the same relation π(x) = ∂0φ(x) holds.

3.4 Quantizing the ABB and ABC-Theory

All of the previous development has been for only one real scalar field. The calculations in Chapter 5 will be for scalar field theories involving more than one particle. So for the ABC theory ([6], [7], [8]) we define the A-field, B-field, and C-field from the following commutation relations

3 [φi(x), πj(y)] = iδ (x − y)δij for i, j = A, B, C (3.54)

[φi(x), φj(y)] = [πi(x), πj(y)] = 0, (3.55)

22 where δij is the Kronecker delta. The ﬁeld operators are deﬁned in an analagous way:

Z d3p 1 φ (x) = a e−ip·x + a† eip·x (3.56) A 3 p p p (2π) 2Ep Z d3p 1 φ (x) = b e−ip·x + b† eip·x (3.57) B 3 p p p (2π) 2Ep Z d3p 1 φ (x) = c e−ip·x + c† eip·x . (3.58) C 3 p p p (2π) 2Ep

The Hamiltonian and momentum are deﬁned similarly to (3.32) and (3.34) by

Z d3p H = E a† a + E b† b + E c† c (3.59) (2π)3 pA p p pB p p pC p p Z d3p P = p a† a + p b† b + p c† c , (3.60) (2π)3 A p p B p p C p p

and a general multi-particle state of l A-particles, m B-particles, and n C-particles can be

written

|pA1 ,..., pAl , pB1 ,..., pBm , pC1 ,..., pCn i

1/2 1/2 1/2 1/2 1/2 1/2 = (2EA1 ) ··· (2EAl ) (2EB1 ) ··· (2EBm ) (2EC1 ) ··· (2ECn )

× a† ··· a† b† ··· b† c† ··· c† . p1 pl p1 pm p1 pn

All these results hold for the ABB-theory ([9]) as well with the obvious removal of the the C-particle and corresponding ﬁeld/creation and annihilation operators. Note also that this is a framework for non-interacting ﬁeld theories. Once we introduce an interaction between the

ﬁelds we may not undertake the same quantization procedure. A perturbative approach must be used to obtain a probability amplitude for a multi-particle state to evolve to another as a power series expansion in g, the coupling between the ﬁelds (the strength of the interaction.)

3.5 Calculating h0|ap, φ(x)|0i

We will need the following result

1 ip·x h0|apφ(x)|0i = p e , (3.61) 2Ep

Note that the aq|0i will cause the ﬁrst term to vanish by deﬁnition of operators acting on the vacuum state. The second term contains a vev seen in previous calculations which can

† be reduced to the commutator [ap, aq]. Finishing the steps, Z d3q 1 Z d3q 1 1 h0|[a , a† ]|0ieiq·x = (2π)3δ3(p − q)eiq·x = eip·x. (3.63) 3 p p q 3 p p (2π) 2Eq (2π) 2Eq 2Ep and applying the rule for integrating with δ-functions, we obtain our desired result (where

p · x = Ept − p · x. A similar result holds for φ(x) and a†

1 −ip·x = p e . 2Ep

24 3.6 Derivation of the Feynman Propagator

The next reasonable calculation would be

h0|φ(x)φ(y)|0i, (3.65) and the following derivation of this quantity is found in [8], sec. 6.3.2. We ﬁrst expand both of the ﬁeld operators into their mode expansions and take the integrals out of the vev:

† † The product in the vev is of the form h0|(ap + ap)(aq + aq)|0i and can be reduced to (2π)3δ3(p − q) in the following way:

−ip·x † ip·x −iq·y † iq·y h0| ape + ape aqe + aqe |0i

−iq·x −iq·y † −ip·x iq·y = h0|apaq|0ie e + h0|apaq|0ie e

† ip·x iq·y † † ip·x iq·y + h0|apaq|0ie e + h0|apaq|0ie e (3.66)

Every term except for the second on the right-hand side vanishes by the rules a|0i = 0 or 0 = h0|a†. The second term itself can be reduced in the usual way to the commutation relation between the two operators:

† 3 3 h0|apaq|0i = (2π) δ (p − q).

It then follows that the vev has been reduced to (2π)3δ3(p − q)e−ip·xeiq·y, so the amplitude becomes ZZ d3p d3q 1 h0|φ(x)φ(y)|0i = 3 3 p (2π) (2π) 2Ep2Eq (3.67) × (2π)3δ3(p − q)e−ip·xeiq·y.

25 We may now choose to evaluate either integral using the δ3(p − q), and here the d3q inte-

gration is performed. We are left with

Z 3 d p 1 −ip(x−y) h0|φ(x)φ(y)|0i = 3 e . (3.68) (2π) 2Ep

Also, to this point we had been ignoring the time ordering of these two ﬁelds

Z 3 d p 1 −ip·(x−y) h0|φ(x)φ(y)|0i = 3 e (2π) 2Ep Z d3p 1 = 3 (2π) 2ωp (3.69)

−iωp(t1−t2) −ip·(x−y) θ(t1 − t2)e e i −iωp(t2−t1) −ip(y−x) + θ(t2 − t1)e e , where we have replaced the Ek with the ωk notation and separated the exponentials into into their respective temporal and spatial coordinates. This form is guaranteed to be covariant by the left-hand side, but the right-hand side does not appear to be so, since the integral is over 3-space and would leave only a function of t which would seem to be non-covariant. However, the key to expressing this integral into a more convincing form is to turn it into a four-fold integral. We do this by introducing an integral representation of θ(t):

Z ∞ dz e−izt θ(t) = i . (3.70) −∞ 2π z + i

Multiplying θ(t) by e−iωt we get

Z ∞ dz e−i(z+ω)t θ(t)e−iωt = i . (3.71) −∞ 2π z + i

And by changing integration variables from z to z + ω, the integral becomes

Z ∞ dz e−izt θ(t)e−iωt = i . (3.72) −∞ 2π z − (ωp − i)

26 To turn the vev into a four-fold integral, we plug this into our original integral

Z d3p h Z ∞ e−iz(t1−t2)+ip·(x−y) h0|T (φ(x)φ(y))|0i = i 4 dz (2π) ωp −∞ z − (ωk + i) Z ∞ e−iz(t2−t1)+ip·(y−x) i + dz , −∞ z − (ωk + i)

and setting z = p0 so that it’s the ﬁrst component of the four-momentum

Z d4p i h e−ip·(x−y) eip·(x−y) i = 4 + . (2π) 2ωp p0 − (ωp − i) p0 + (ωp − i (3.73)

Finally, by changing the integration variable in the second integral from p to −p, we obtain a common denominator so the exponential may be taken out and we get

Z 4 d p ip·(x−y) i h0|T {φ(x)φ(y)})|0i = 4 e 2 2 . (3.74) (2π) (p0) − (ωp − i)

The last step is to simplify the denominator of the fraction in the integrand by noting recalling (3.7) so that

2 2 2 2 2 p0 − (ωp − i) = p0 − ωp + 2ωpi − (i)

2 2 = p0 − ωp + 2ωpi

2 2 = p0 − ωp + i

2 2 2 = p0 − p − m + i,

2 where we dropped the term of order O( ) in the second step, absorbed the 2 and ωp (in

which we assumed ωp > 0). The ﬁnal expression for the propagator is expressed as follows:

Z 4 d p ip·(x−y) i h0|T (φ(x)φ(y)|0i = 4 e 2 2 2 . (3.75) (2π) p0 − p − m + i This quantity is known as the Feynman Propagator and in the literature it is commonly denoted as D(x − y) or ∆(x − y). This is obviously an expression in position space, and one can clearly see that the expression in momentum space is just

i D(p) = . (3.76) p2 − m2 + i

27 In either representation it has the interpretation of a particle propagating from spacetime points x to y. Mathematically, it is the Green’s function of the Klein-Gordon equation in momentum space.

28 CHAPTER 4

PERTURBATION THEORY

We have previously described free, non-interacting scalar multi-particle states as elements |p1, p2,..., pni of the constructed Fock space. We are now in a position to describe how states evolve from one to another. Therefore we attempt to solve quantum ﬁeld theories of the form

H = H0 + Hint, (4.1) where H0 is the Hamiltonian corresponding to a (classical) free scalar ﬁeld theory and Hint an an interacting Hamiltonian with a corresponding density Hint deﬁned in the usual way by Z Z 3 3 Hint = d xHint = − d xLint. (4.2)

Note that Hint = −Lint.

4.1 Deriving the perturbation expansion

We must resort instead to a perturbative approach which will allow us to use our interacting fields and multiparticle states in terms of the free fields above and the multi- particle states in the product Fock space. Following [4] (sec. 4.2), we will derive an expression for the two-point correlation function in φ3 theory, a theory describing one scalar field with a cubic interaction λ L = − φ3. (4.3) int 3! Denote |Ωi as the ground state of the interacting theory (|0i is still the ground state of the free theory). We want to compute the quantity

hΩ|T φ(x)φ(y)|Ωi, (4.4) where φ in this case is the ﬁeld corresponding to the interacting theory. The goal here is to express both |Ωi and φ, the interaction theory ground state and ﬁeld, in terms of |0i and

φI (x), the free theory ground state and ﬁeld.

29 4.2 Expressing φ(x) in Terms of φI (x)

First we note that in the Heisenberg picture the full Heisenberg ﬁeld evolves in time according to φ(x) = e−iHtφ(x)eiHt. (4.5)

Now at any time t0 the ﬁeld may written in terms of its creation and annihilation operators

Z d3 1 φ(t , x) = a eip·x + a† e−ip·x , (4.6) 0 3 p p p (2π) 2Ep

then for any subsequent time t the ﬁeld evolves according to

−iH(t−t0) iH(t−t0) φ(t, x) = e φ(t0, x)e . (4.7)

If there was no interaction term, no coupling between the fields, then the above evolution would be for a free field. It commonly referred to as φI (x), the interaction picture field

Z d3p 1 φ (x) = a eip·x + a† e−ip·x , (4.8) I 3 p p p (2π) ) 2Ep

0 where x = t − t0. We can then express the full Hamiltonian ﬁeld φ(x) from this interaction

picture ﬁeld by introducing the time-evolution operator U(t, t0) deﬁned by

iH0t −iH(t−t0) U(t, t0) ≡ e e . (4.9)

Then φ(x) can be expressed as

† φ(x) = U (t, t0)φI (x)U(t, t0)

iH(t−t0) −iH0t iH0t −iH(t−t0) = e e φI (x)e e iH(t−t0) −iH0t iH0t −iH0t iH0t −iH(t−t0) = e e e φ(t0, x)e e e

iH(t−t0) −iH(t−t0) = e φ(t0, x)e .

Now U(t, t0) satisﬁes the Schrodinger equation

∂ i U(t, t ) = H (t)U(t, t ), (4.10) ∂t 0 I 0

30 iH0(t−t0) −iH0(t−t0) where HI (t) = e (Hint)e . Clearly the solution to (4.10) must be some form

of exponential, and the true solution takes the form of a power series in HI (t)

Z t Z t Z t1 2 U(t, t0) = 1 + (−i) dt1HI (t1) + (−i) dt1 dt2 HI (t1)HI (t2) (4.11) t0 t0 t0 Z t Z t1 Z t2 3 + (−i) dt1 dt2 dt3HI (t1)HI (t2)HI (t3) + ··· . (4.12) t0 t0 t0 Note that for each integral in the sum we have

Z t Z t1 Z tn−1 1 Z t dt1 dt2 ··· dtnH(t1) ··· H(tn) = dt1 ··· dtnT {HI (t1) ··· HI (tn)}, (4.13) t0 t0 t0 n! t0 thus we are left with Z t n h 0 0 io U(t, t0) ≡ T exp − i dt HI (t ) . (4.14) t0 and we arrive at our result of expressing the full Hamiltonian ﬁeld φ(x) in terms of the free

ﬁeld φI (x). Combining this result with the expression of |Ωi in terms of |0i by evolving |0i

with the time-evolution operator U(T, t0) (see [4] section 4.2)

U(T, t )|0i |Ωi = lim 0 (4.15) T →∞(1−i) e−iE0(t0−T )hΩ|0i

for which hΩ|0i was assumed nonzero and E0 is the dominent term in the evolution as T → ∞; we are left with our ﬁnal result ([4], eq. 4.31)

n h R t 0 0 io h0|T φI (x)φI (y) exp − i −t dt HI (t ) |0i hΩ|T {φ(x)φ(y)}|Ωi = lim n h io . (4.16) t→∞(1−i) R t 0 0 h0|T exp − i −t dt HI (t ) |0i From this expression we can calculate S-matrix elements, or scattering amplitudes, for different processes. Given some initial state |ii and a final state |fi, the probability of a state evolving after some period of time to the final state is given by

hf|S|ii, (4.17)

where the S-operator is given by the Dyson expansion (see eq. 6.42 of [8])

∞ X (−i)n Z S = d4x d4x ··· d4x H (x )H (x ) ··· H (x ). (4.18) n! 1 2 n I 1 I 2 I n 0

31 CHAPTER 5

TREE-LEVEL AMPLITUDES IN ABC AND ABB THEORY

From the Dyson expansion ([6], [7], [9], [8]) we can immediately start calculating

S-matrix elements for diﬀerent theories.

5.1 AB-Scattering in ABC Theory

Consider two incoming A and B particles with momentum pA and pB in ABC-theory. The two-particle state representing this is the incoming “ket”

1/2 1/2 † † |pA, pBi = (2EpA ) (2EpB ) apbp|0i. (5.1)

Upon an interaction we are left two outgoing A and B particles with momentum qA and qB which are represented by the outgoing “bra”

1/2 1/2 hqA, qB| = h0|aqbq(2EqA ) (2EqB ) . (5.2)

We calculate up to and including second order the terms in the perturbative expansion of

the S-matrix element hqA, qB|S||pA, pBi in the ABC theory (HI = gφAφBφC ).

5.1.1 The Zeroth Order Term in Expansion of g

We ﬁrst calculate the 0-th order term hqA, qB|I||pA, pBi. Performing the calculation we get

1/2 † † hqA, qB|I|pA, pBi = (16EqA EqB EpB EpA ) h0|aqbqapbp|0i.

† We interchange the bq and ap and subtract a term with the two a operators ﬂipped to obtain a commutator in the vev

† † † † h0|aqapbqbp|0i = h0|[aq, ap]bqbp|0i. (5.3)

This commutator can be taken out of the vev and will enforce qA = pA. We repeat the

process for the b operators to obtain another commutator which enforces qB = pB. The

32 δ-functions also enforce conservation of energy of each particle from the convervation of

momentum as follows:

q q 2 2 2 2 EqA = (qA) + mA = (pA) + mA = EpA (5.4)

and similarly for EB. The resulting 0-th order term is

1/2 1/2 3 3 (2EA) (2EB) δ (qA − pA)δ (qB − pB). (5.5)

This calculation shows that the 0-th order term in the expansion is merely the two particles passing by each other not interacting as shown in the following ﬁgure.

p A A

p B B

Figure 5.1: Zeroth-order term in the power series expansion of g

5.1.2 The First Order Term in Expansion of g

Next we proceed to calculate the ﬁrst order term in the series expansion. It is the

ﬁrst term to include the actual interaction, and it is of the form

Z 4 − ig d xhqA, qB|φAφBφC |pA, pBi. (5.6)

Recall that each φi is an operator in the interaction picture; thus each φi is expressed by a standard mode expansion of free creation and annihilation operators:

Z d3q 1 φ(x) = a e−iq·x + a† eiq·x . (5.7) 3 p q q (2π) 2Eq

33 We expand the ﬁrst term in the amplitude as follows:

Z Z 4 4 1/2 1/2 −ig d xhqA, qB|φAφBφC |pA, pBi = −ig d xh0|aqbq(2EqA ) (2EqB ) Z 3 Z 3 d kA 1 d kB 1 a e−ikA·x + a† eikA·x b e−ikB ·x + b† eikB ·x (2π)3 p k k (2π)3 p k k 2EkA 2EkB Z 3 d kC 1 c e−ikC ·x + c† eikC ·x (2E )1/2(2E )1/2a† b† |0i. (2π)3 p k k pA pB p p 2Ekc

We rearrange the terms and combine the integrals so that only the creation and annihilation operators (with associated exponentials where appropriate) are present within the vacuum expectation

Z 3 3 3 1/2 4 d kA d kB d kC 2EqA 2EqB 2EpA 2EpB −ig d x 3 3 3 (2π) (2π) (2π) 2EkA 2EkB 2EkC

−ikA·x † ikA·x −ikB ·x † ikB ·x −ikC ·x † ikC ·x † † h0|aqbq(ake + ake )(bke + bke )(cke + cke )apbp|0i.

Suppressing the associated exponentials, the vacuum expectation value (vev) takes the following form:

† † † † † h0|aqbq(ak + ak)(bk + bk)(ck + ck)apbp|0i. (5.8)

Notice that by the commutation relations imposed at the outset (3 noninteracting free ﬁelds)

† we can move either the ck or the ck to the appropriate h0| or |0i and the entire expectation vanishes;

† † † † † † † † † h0|aqbq(ak + ak)(bk + bk)ckapbp|0i + h0|aqbq(ak + ak)(bk + bk)ckapbp|0i

† † † † † † † † † † = h0|aqbq(ak + ak)(bk + bk)apbpck|0i + h0|ckaqbq(ak + ak)(bk + bk)ckapbp|0i

= 0.

Since the vacuum expectation value vanishes, the entire integral goes to 0, thus the ﬁrst-

order term in the expansion is zero. Note that there is no Feynman diagram for ﬁrst order in this case.

34 5.1.3 The Second-Order Term in Expansion of g

Next we calculate the second-order term in the perturbative expansion. It takes the

form (−ig)2 ZZ d4xd4yhq , q |T {φ φ φ φ φ φ }|p , p i. (5.9) 2 A B Ax Bx Cx Ay By Cy A B Rather than expressing each ﬁeld into its respective mode expansion of free creation and annihilation operators and rearranging terms to cancel members of the sum of products of

operators we defer to some algebraic techniques to simplify the product. The discussion follows [8], but the identity below is a special case of the lemma in Chapter 2. Consider the vacuum expectation h0|ABCD|0i where A, B, C, and D are arbitrary

creation operators, annihilation operators, or linear combinations of the two. If A = a + a†, then

h0|ABCD|0i = h0|aBCD|0i + h0|a†BCD|0i

= h0|aBCD|0i − h0|BCDa|0i + h0|a†BCD|0i

= h0|aBCD − BCDa|0i + 0

= h0|[a, BCD]|0i.

Using the simple identity

[a, BCD] = [a, B]CD + B[a, C]D + BC[a, D] (5.10)

given by the proof

[a, BCD] = aBCD − BCDa

= aBCD − BaCD + BaCD − BCDa

= (aB − Ba)CD + BaCD − BCaD + BCaD − BCDa

= [a, B]CD + B(aC − Ca)D + BC(aD − Da)

= [a, B]CD + B[a, C]D + BC[a, D],

35 the vev becomes

h0|[a, BCD]|0i = [a, B]h0|CD|0i + [a, C]h0|BD|0i + [a, D]h0|BC|0i. (5.11)

Note that for each of the commutators above which are in the form [a, X], the following substitution may be made (since h0|0i)

[a, X] = h0|[a, X]|0i = h0|aX|0i − h0|Xa|0i = h0|aX|0i. (5.12)

Note also that [X, a†] = h0|Xa†|0i. Plugging (5.12) into the vev we get

h0|ABCD|0i =h0|ab|0ih0|CD|0i + h0|aC|0ih0|BD|0i

+ h0|aD|0ih0|BC|0i.

Thus the vev of a product of multiple creation and annihilation operators has been reduced to sums of products of vev’s of only two. Typically most of these terms will vanish by the rules set forth in the beginning for creation and annihilation operators acting on vacuum states. This result holds much more generally to time ordered products of ﬁelds. It is known as Wick’s Theorem and is used in advanced treatments on QFT (e.g., [4], [5]) as a device to reduce a large product of ﬁelds, say in an interaction term, vev’s of commutators which ultimately become Feynman propagators. The matrix element takes the form

ZZ 2 0 0 (−ig) x1x2hpA, pB|T {φA(x1)φB(x1)φC (x1)φA(x2)φB(x2)φC (x2)}|pA, pBi. (5.13)

There are two diagrams which correspond to a nonzero amplitude; these will be discussed in the following sections.

0 2 u-Channel Process: u = (pA − pB)

The ﬁrst way that we say the particles can interact is by exchanging a C-particle as indicated in the ﬁgure. An A and a B-particle are accelerated to relativistic energies, and upon a short-range interaction, they exchange a virtual C-particle. This is known as a

36 A B

p q A B

k C C

p q B A

B A

0 Figure 5.2: C-Exchange in AB-scattering. The qA refers to the pA in the discussion and similarly for qB.

u-channel process, and the integral describing the process is ZZ 2 † † (−ig) dx1dx2h0|φA(x2)ap|0ih0|φB(x1)bp|0i

p 0 0 h0|φC (x1)φC (x2)|0ih0|ap0 φA(x1)|0ih0|bp0 φB(x2)|0i 16EAEBEAEB. (5.14)

From the properties calculated in the previous section we can replace the vev’s with the

corresponding exponential or propagating term as follows:

ZZ 1 1 1 0 1 0 2 −ipA·x1 −ipB ·x2 ipA·x2 ipB ·x1 (−ig) dx1dx2 e e 0 e 0 e 2EA 2EB 2EA 2EB Z 4 (5.15) d k −ik·(x −x )p e 1 2 16E0 E0 E E . (2π)4 A B A B

Cancelling the normalization factors and gathering all the exponentials not in the propagator

term we get

ZZ Z 4 2 4 x i(p0B−p )·x i(p0 −p )·x d k −ik·(x −x ) (−ig) d x d4 e A 1 e A B 2 e 1 2 . (5.16) 1 2 (2π)4

4 4 We will now evaluate the d x1 and d x2 integrals. Notice that the propagator term is

a function of the diﬀerence x1 − x2 of the coordinates. We expect this to be so, because the physics of the situation should be independant of the choice of the coordinate system [8]. To

37 show this explicity in the calculation we change the variable of the space integrations from x1 and x2 to x and X which are deﬁned coordinate-wise by

x + x x = x − x X = 1 2 . (5.17) 1 2 2

It is useful at this point to make a remark. Were the integration variables one-dimensional over Euclidean space, the change of variables formula would take the form

ZZ ZZ ∂(x , x ) dx dx f(x , x ) = dxdXf(x (x, X)x (x, X)) 1 2 , (5.18) 1 2 1 2 1 2 ∂(x, X) where ∂(x1,x2) is the determinant of the Jacobian matrix in the change of variables formula ∂(x,X) (see [12], sec. 6.2, p. 377). The simple calculation would show that this is unity:

∂x1 ∂x1 1 ∂(x1, x2) ∂x ∂X 2 1 = = = 1. (5.19) ∂(x, X) ∂x2 ∂x2 −1 ∂x ∂X 2 1 However we are changing two integration variables in a 4-dimensional Minkowski space, so we are actually changing from a set of 8 one-dimensional variables to another set of 8. The actual Jacobian in the change of variables looks like

 0 0 0 0 0 0 0 0  ∂x1 ∂x1 ∂x1 ∂x1 ∂x1 ∂x1 ∂x1 ∂x1 ∂x0 ∂x1 ∂x2 ∂x3 ∂X0 ∂X1 ∂x1 ∂X3  ∂x1 ∂x1 ∂x1 ∂x1 ∂x1 ∂x1 ∂x1 ∂x1   1 1 1 1 1 1 1 1   ∂x0 ∂x1 ∂x3 ∂x3 ∂X0 ∂X1 ∂X2 ∂X3   ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2   1 1 1 1 1 1 1 1   ∂x0 ∂x1 ∂x3 ∂x3 ∂X0 ∂X1 ∂X2 ∂X3   ∂x3 ∂x3 ∂x3 ∂x3 ∂x3 ∂x3 ∂x3 ∂x3   1 1 1 1 1 1 1 1   ∂x0 ∂x1 ∂x2 ∂x3 ∂X0 ∂X1 ∂X2 ∂X3   0 0 0 0 0 0 0 0  . (5.20)  ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2   ∂x0 ∂x1 ∂x2 ∂x3 ∂X0 ∂X1 ∂X2 ∂X3   1 1 1 1 1 1 1 1   ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2   ∂x0 ∂x1 ∂x2 ∂x3 ∂X0 ∂X1 ∂X2 ∂X3   2 2 2 2 2 2 2 2   ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2   ∂x0 ∂x1 ∂x2 ∂x3 ∂X0 ∂X1 ∂X2 ∂X3   3 3 3 3 3 3 3 3  ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x2 ∂x0 ∂x1 ∂x2 ∂x3 ∂X0 ∂X1 ∂X2 ∂X3

The corresponding determinant is then calculated by taking each partial derivative for each

38 coordinate:

1 0 0 0 1 0 0 0 2

1 0 2 0 0 0 1 0 0

1 0 0 2 0 0 0 1 0

1 0 0 0 2 0 0 0 1 = 1. (5.21) 1 − 2 0 0 0 1 0 0 0

1 0 − 2 0 0 0 1 0 0

1 0 0 − 2 0 0 0 1 0

1 0 0 0 − 2 0 0 0 1

Once again, we obtain unity for the proportionality between the dx1dx2 variables and the dxdX variables.

Also we need to solve (5.17) for x1 and x2 which can be done by the simple substitution method of linear algebra: 1 −1 x = x + X x = x + X. (5.22) 1 2 2 2 With these two we can ﬁnally make the change of variables in (5.16) to get

2 ZZ x −x −ig i(p0 −p )·( +X) i(p0 −p )·( +X) d4xd4Xe B A 2 e A B 2 2 Z 4 (5.23) d kC −ik·x i 4 e 2 2 . (2π) kC − mC + i Multiplying terms in the exponential and rearranging them to calculate the dX integral ﬁrst

we obtain 2 Z Z x −ig X i(p0 −p +p0 −p )·X 4 i(p0 −p −p0 +p )· d e B A A B d xe B A A B 2 2 Z 4 (5.24) d kC −ik·x i × 4 e 2 2 . (2π) kC − mC + i The dX integral is clearly now calculated to be a δ-function. We can use the even function property of the δ-function to switch the signs on the pi’s inside: −ig 2 (2π)4δ4(p + p − p0 − p0 ) 2 A B A B Z x 4 i(p0 −p −p0 +p )· × d xe B A A B 2 (5.25) Z 4 d k −ik·x i × 4 e 2 2 . (2π) kC − mC + i

39 0 0 Notice that the δ-function implies that conservation of momentum: pA + pB = pA + pB. We

0 can then deﬁne the four-momentum transfer q by q = pB − pA. By the δ-function, this also

0 enforces q = pB − pA so that the term in the exponential of the dx integral becomes

0 0 x x i(pB − pA − pA + pB) · 2 = i(q + q) · 2 = i(q · x). (5.26)

Plugging this new expression in and switching the integrals (with appropriate continuity assumptions), we get

−ig 2 (2π)4δ4(p + p − p0 − p0 ) 2 A B A B (5.27) Z d4k h Z i i 4 i(q−kC )·x × 4 d xe 2 2 . (2π) kC − mC + i

4 4 The bracketed integral is of course (2π) δ (q − kC ) which when used in the dkC integration

4 singles out q as a particular value of kC which leaves us with the ﬁnal result (the (2π) ’s obviously cancel)

2 −ig 4 4 0 0 i (2π) δ (pA + pB − pA − pB) 2 2 . (5.28) 2 q − mC + i

This entire process can be repeated by switching the x1 and x2 in the original expression of the matrix element. The exact same result as (5.28) would be obtained and added to it, cancelling the 2’s. Doing this we obtain as our ﬁnal result

2 4 4 0 0 i M = (−ig) (2π) δ (pA + pB − pA − pB) 0 2 2 . (5.29) (pB − pA) − mC + i

Once again this is a u-channel process (see [8], sec. 6.3, eq. 6.104.) Any time a tree-level scattering process is considered, it will occur through diﬀerent channels. The second possible channel is discussed next.

2 s-Channel Process: s = (pA + pB)

The second non-zero contribution to the scattering amplitude (5.13) is shown in the following ﬁgure. In this process the incoming particles interact through a temporary annihilation reproduction of one another through a virtual C-particle. For the expression to

40 p q A A A A

k C p q B B B B

0 Figure 5.3: AB Annihilation for ABC Theory. qA corresponds to pA and similarly for qB.

be computed we have

−ig 2 ZZ d4x d4x p16E0 E0 E E 2 1 2 A B A B

h0|ap0 φA(x2)|0ih0|bp0 φB(x2)|0i (5.30)

h0|φC (x1)φC (x2)|0i

† † h0|φA(x1)ap|0ih0|φB(x2)bp|0i. We calculate this integral in a similar manner as the u-channel process. Reorganizing terms and subtituting values we get

−ig 2 ZZ 1 0 1 0 4 4 p 0 0 ip ·x2 ip ·x2 d x1d x2 16E E EAEB e A e B 2 A B p2E0 p2E0 A B (5.31) Z 4 d kC i 1 1 −ik·x √ −ipA·x1 √ −ipB ·x1 4 e 2 2 e e . (2π) kC − mC + i 2EA 2EB Cancelling the normalizations and gathering like terms we further obtain

2 ZZ Z 4 −ig 0 0 d kC i 4 4 −i(pA+pB )·x1 i(pA+pB )·x2 −ik·x d x1d x2e e 4 e 2 2 . (5.32) 2 (2π) kC − mC + i

Once again we make the change of variables (with the same Jacobian)

x = x1 − x2 x1 = X + x/2

X = (x1 + x2)/2 x2 = X − x/2.

41 Upon substitution, the exponentials in the x1x2 integral become

−i(p +p )·(X+x/2) i(p0 +p0 )·(X−x/2) −i(p +p )·X −i(p +p )·x/2 i(p0 +p0 )·X −i(p0 +p0 )·x/2 e A B e A B = e A B e A B e A B e A B

i(p0 +p0 −p −p )·X −i(p +p +p0 +p0 )·x/2) = e A B A B e A B A B .

We may thus rewrite (5.32) as

2 Z Z −ig 4 i(p0 +p0 −p −p )·X 4 −i(p +p +p0 +p0 )·x/2) d Xe A B A B d xe A B A B 2 Z 4 (5.33) d kC −ik·x i 4 e 2 2 . (2π) kC − mC + i The dX integral becomes a δ-function

Z 0 0 4 i(pA+pB −pA−pB )·X 4 4 0 0 d Xe = (2π) δ (pA + pB − pA − pB),

0 0 and the δ-function enforces pA + pB = pA + pB which cancels the 2 in the exponential of the dx integral. Combining this and interchanging the order of the dx and dk integrals we obtain

−ig 2 Z d4k h Z i i 4 4 0 0 C 4 i(pA+pB −kC )·x) (2π) δ (pA + pB − pA − pB) 4 d xe 2 2 . (5.34) 2 (2π) kC − mC + i

The dx integral is another δ-function

Z 4 i(pA+pB −kC )·x) 4 4 d xe = (2π) δ (pA + pB − kC ), (5.35)

so with this result we can calculate the ﬁnal integral

2 Z 4 −ig 4 4 0 0 d kC 4 4 i (2π) δ (pA + pB − pA − pB) 4 (2π) δ (pA + pB − kC ) 2 2 (5.36) 2 (2π) kC − mC + i which gives our ﬁnal result (note that we obtain an identical result by interchanging the x1

and x2 values which cancels the two in front)

2 4 4 0 0 i (−ig) (2π) δ (pA + pB − pA − pB) 2 2 . (5.37) (pA + pB) − mC + i

42 B p B

p C C

p A A

Figure 5.4: A C-particle decaying into an A and a B-particle

5.2 C → A + B Decay

The next process we calculate is C-decay into two particles: an A and a B. The corresponding ﬁgure below illustrates the process. The amplitude we seek to calculate is

Z 4 (−ig)hpb, pc| d xφAφBφC |pci. (5.38)

We expand this term using the mode expansion of ﬁelds and the momentum states constructed from the vacuum,

Z 4 (−ig)hpB, pA| d xφAφBφC |pC i (5.39) Z Z 3 p p d kA 1 † = −igh0|a b 2E 2E d4x (a e−ikA·x + a )eikA·x) p p A B (2π)3 p k k 2EkA Z 3 Z 3 d kB 1 d kC 1 × (b e−ikB ·x + b† eikB ·x) (c e−ikC ·x + c† eikC ·x) (2π)3 p k k (2π)3 p k k 2EkB 2EkC p † × 2EC cp|0i. (5.40)

Rearranging terms so that only the creation and annihilation operators (and associated exponentials) are contained within the vev we obtain:

Z 3 3 3 s 4 d kA d kB d kC 2EA2EB2EC −ig d x 3 3 3 (2π) (2π) (2π) 2EkA 2EkB 2EkC

−ikA·x † ikA·x −ikB ·x † ikB ·x h0|apbp(ake + ake )(bke + bke )

−ikC ·x † ikC ·x † (cke + cke )cp|0i.

43 Suppressing the exponentials, the vev in the integral above takes the following form

† † † † h0|apbp(ak + ak)(bk + bk)(ck + ck)cp|0i.

We systematically reduce this sum of 8 products of 6 creation/annihilation operators to a single product of operators which will in turn be used with commutation relations to

† transform the vev to a product of δ-functions. To begin the process we ﬁrst eliminate the ck by moving it to the left

† † † † h0|apbp(ak + ak)(bk + bk)(ck + ck)cp|0i

† † † † † † † = h0|apbp(ak + ak)(bk + bk)ckcp|0i + h0|apbp(ak + ak)(bk + bk)ckcp|0i

† † † † † † † = h0|apbp(ak + ak)(bk + bk)ckcp|0i + h0|ckapbp(ak + ak)(bk + bk)cp|0i

† † † = h0|apbp(ak + ak)(bk + bk)ckcp|0i + 0,

then moving bk to the right

† † † † † = h0|apbp(ak + ak)bkckcp|0i + h0|apbp(ak + ak)bkckcp|0i

† † † † † = h0|apbp(ak + ak)ckcpbk|0i + h0|apbp(ak + ak)bkckcp|0i

† † † = 0 + h0|apbp(ak + ak)bkckcp|0i,

and ﬁnally moving ak to the right in the same way

† † † † † = h0|apbpakbkckcp|0i + h0|apbpakbkckcp|0i

† † † † † = h0|apbpbkckcpak|0i + h0|apbpakbkckcp|0i

† † † = 0 + h0|apbpakbkckcp|0i.

We obtain a simpliﬁcation of the original vev

† † † † † † † h0|apbp(ak + ak)(bk + bk)(ck + ck)cp|0i = h0|apbpakbkckcp|0i. (5.41)

From this reduction we then use each product aa† to turn the vev into a product of δ-functions using the commutation relations given by (3.12). The calculation proceeds as

44 follows. First we do the commutator for A;

† † † † † † h0|apbpakbkckcp|0i = h0|apakbpbkckcp|0i

† † † † = h0|(apak − akak)bpbkckcp|0i

† † † = h0|[ap, ak]bpbkckcp|0i

3 3 † † = (2π) δ (pA − kA)h0|bpbkckcp|0i.

Then we perform the commutator for B;

3 3 † † † = (2π) δ (pA − kA)h0|(bpbk − bkbk)ckcp|0i

3 3 † † = (2π) δ (pA − kA)h0|[bp, bk]ckcp|0i

3 3 3 3 † = (2π) δ (pA − kA)(2π) δ (pB − kB)h0|ckcp|0i.

Finally we perform the commutator for C;

3 3 3 3 † † = (2π) δ (pA − kA)(2π) δ (pB − kB)h0|ckcp − cpck|0i

3 3 3 3 † = (2π) δ (pA − kA)(2π) δ (pB − kB)h0|[ck, cp]|0i

3 3 3 3 3 3 = (2π) δ (pA − kA)(2π) δ (pB − kB)(2π) δ (kC − pC )h0|0i

3 3 3 3 3 3 = (2π) δ (pA − kA)(2π) δ (pB − kB)(2π) δ (kC − pC ).

Thus the vev of the product of operators has been reduced to a product ofδ-functions

† † † 3 3 3 3 3 3 h0|apbpakbkckcp|0i = (2π) δ (pA − kA)(2π) δ (pB − kB)(2π) δ (kC − pC ). (5.42)

We can now take this expression for the vev and plug it back into our integral,

ikA·x ikB ·x −ikC ·x † † remebering to include the appropriate exponentials: e , e , and e for ak, bk, and ck, respectively.

Z 3 3 3 s 4 d kA d kB d kB 2EA2EB2EC M = (−ig) d x 3 3 3 (2π) (2π) (2π) 2EkA 2EkB 2EkC

3 3 3 3 3 3 i(kA+kB −kC )·x × (2π) δ (pA − kA)(2π) δ (pB − kB)(2π) δ (kC − pC )e .

45 After the simpliﬁcation of the vev we are in a position to actually calculate the integral.

Using the property of δ-functions on the integrals over k-space Z d3k f(k)δ3(p − k) = f(p) (5.43) (2π)3 and then taking the Fourier Transform of the resulting exponential in 4-space we obtain after rearranging terms s Z Z d3k 2E 4 A A ikA·x 3 3 M = (−ig) d x 3 e (2π) δ (pA − kA) (2π) 2EkA s s Z d3k 2E Z d3k 2E B B ikB ·x 3 3 C C −ikC ·x 3 3 × 3 e (2π) δ (pB − kB) 3 e (2π) δ (kC − pC ) (2π) 2EkB (2π) 2EkC Z = (−ig) d4xeipA·xeipB ·xe−ipC ·x Z = (−ig) d4xei(pA+pB −pC )·x

4 4 = (−ig)(2π) δ (pA + pB − pC ).

We have at last obtained the amplitude for C-decay into an A and a B particle

4 4 M = (−ig)(2π) δ (pA + pB − pC ). (5.44)

At this point it is worthwhile to make a remark. Although this is a scattering amplitude for a C-particle into an A and a B, what we in eﬀect have just done is derive a Feynman rule of ABC theory. For any diagram that should occur in a process described by an ABC

interaction, every vertex will look like the one in Figure 5.4. We can simply write (−ig) for each vertex instead of repeating this entire calculation. Every Feynman diagram with vertices will include this factor as many times as necessary.

Another remark is that (−ig) is part of the invariant amplitude. Diﬀerent texts give

4 4 different arguments for why the (2π) δ (pA +pB −pC ) can be removed. [4] solves the problem formally using wave packets, but [13] uses the quicker method of letting the particles interact in a finite volume for a finite time so that an element of the S-matrix may ultimately written as

4 S = −2πiδ (pA + pB − pC )M. (5.45)

46 5.3 A → B + B Decay

We now calculate the decay amplitude for the process of an A-particle decaying into two B particles in ABB-theory. The calculation is similar to C → A + B in ABC-theory, but a couple new features arise due to the indistinguishablity of the two resulting particles.

We will show that we recover the same amplitude as for C-decay (interpreting the g as the coupling for ABB theory). The ﬁgure has similar structure to the ABC-vertex and is shown below.

p 1

p A A

p 2

Figure 5.5: An A-particle decaying into 2 B-particles

We proceed as before by ﬁrst writing the S-matrix element out with all the terms explicitly stated. For this calculation we will denote the operators and ﬁelds corresponding to the B-particles with a 1 or 2 subscript. The A-particle is denoted as usual.

(−ig) Z M = hp , p | d4xφ (x)φ (x)φ (x)|p i 2 1 2 1 2 A A Z Z 3 (−ig) p p d k1 1 † = d4xh0|b b 2E 2E (b e−ik1·x + b eik1·x) 2 p1 p2 1 2 (2π)3 p 1 1 2Ek1 Z 3 Z 3 d k2 1 d kA 1 × (b e−ik2·x + b†eik2·x) (a e−ikA·x + a† eikA·x) (2π)3 p 2 2 (2π)3 p k k 2Ek2 2EkA p † × 2EAap|0i.

From this expansion we rearrange terms so that only the creation and annihilation operators

47 (with appropriate exponentials) are left within the vev;

Z 3 3 3 s −ig 4 d k1 d k2 d kA 2E12E22Ea M = d x 3 3 3 2 (2π) (2π) (2π) 2Ek1 2Ek2 2EkA

−ik1·x † ik1·x −ik2·x † ik2·x −ikA·x † ikA·x † × h0|bp1 bp2 (b1e + b1e )(b2e + b2e )(ake + ake )ap|0i.

Suppressing the exponentials, the vev takes the following form:

† † † † h0|bp1 bp2 (b1 + b1)(b2 + b2)(ak + ak)ap|0i. (5.46)

Once again we are tasked with reducing this vev to a product of δ-functions. We follow the

† same manipulations to simplify it. First we distribute the ak and ak;

† † † † h0|bp1 bp2 (b1 + b1)(b2 + b2)(ak + ak)ap|0i

† † † † † † † = h0|bp1 bp2 (b1 + b1)(b2 + b2)akap|0i + h0|bp1 bp2 (b1 + b1)(b2 + b2)akap|0i

† † † † † † † = h0|bp1 bp2 (b1 + b1)(b2 + b2)akap|0i + h0|akbp1 bp2 (b1 + b1)(b2 + b2)ap|0i

† † † = h0|bp1 bp2 (b1 + b1)(b2 + b2)akap|0i + 0.

Then we distribute the b and b† ; k2 k2

† † † † † = h0|bp1 bp2 (b1 + b1)b2akap|0i + h0|bp1 bp2 (b1 + b1)b2akap|0i

† † † † † = h0|bp1 bp2 (b1 + b1)akapb2|0i + h0|bp1 bp2 (b1 + b1)b2akap|0i

† † † = 0 + h0|bp1 bp2 (b1 + b1)b2akap|0i.

We have reduced the vev to the following simpler form

† † † h0|bp1 bp2 (b1 + b1)b2akap|0i. (5.47)

† We can eliminate the product akap just as we did in the C-decay of ABC theory: simply

† subtract a vev with the term apak to produce a commutator. It proceeds as follows;

† † † † † † † h0|bp1 bp2 (b1 + b1)b2akap|0i = h0|bp1 bp2 (b1 + b1)b2(akap − apak)|0i

† † † = h0|bp1 bp2 (b1 + b1)b2[ak, ap]|0i (5.48)

3 3 † † = (2π) δ (pA − kA)h0|bp1 bp2 (b1 + b1)b2|0i.

48 To deal with this reduced vev,

† † † † † h0|bp1 bp2 (b1 + b1)b2|0i = h0|bp1 bp2 b1b2|0i + h0|bp1 bp2 b1b2|0i, (5.49)

† 3 3 † we ﬁrst insert the relation b1b2 = (2π) δ (k1 − k2) + b2b1 into the ﬁrst vev on the right hand side to get

† 3 3 † h0|bp1 bp2 b1b2|0i = h0|bp1 bp2 ((2π) δ (k1 − k2) + b2b1)|0i

3 3 † = (2π) δ (k1 − k2)h0|bp1 bp2 |0i + h0|bp1 bp2 b2b1|0i

= 0.

For the second term on the right-hand side of we apply the same method twice to ultimately

† obtain a product of δ-functions; ﬁrst we do the 2 middle terms bp2 and b1

† † 3 3 † † h0|bp1 bp2 b1b2|0i = h0|bp1 ((2π) δ (p2 − k1) + b1bp2 )b2|0i

3 3 † † † = (2π) δ (p2 − k1)h0|bp1 b2|0i + h0|bp1 b1bp2 b2|0i.

† Doing the same thing for bp1 b2 in the ﬁrst term,

† † 3 3 3 3 † † † h0|bp1 bp2 b1b2|0i = (2π) δ (p2 − k1)h0|(2π) δ (p1 − k2) + b2bp1 |0i + h0|bp1 b1bp2 b2|0i

3 3 3 3 † † = (2π) δ (p2 − k1)(2π) δ (p1 − k2) + h0|bp1 b1bp2 b2|0i.

† † † For the term h0|bp1 b1bp2 b2|0i we use a similar technique as in the case of reducing the aa to reduce it to another product of δ-functions

† † † † † h0|bp1 b1bp2 b2|0i = h0|(bp1 b1 − b1bp1 )bp2 b2|0i

† † = h0|[bp1 , b1]bp2 b2|0i

3 3 † = (2π) δ (p1 − k1)h0|bp2 b2|0i,

† and doing the same thing for bp2 b2,

3 3 † † = (2π) δ (p1 − k1)h0|bp2 b2 − b2bp2 |0i

3 3 3 3 = (2π) δ (p1 − k1)(2π) δ (p2 − k2).

49 Thus our initial vev has been reduce to the following product of δ-functions:

3 3 3 3 3 3 3 3 (2π) δ (p2 − k1)(2π) δ (p1 − k2) + (2π) δ (p1 − k1)(2π) δ (p2 − k2).

Notice that there are two products instead of one as was the case in calculating C → A + B in the ABC theory. Since the ABB theory contains 2 indistinguishable particles, we need

1 1 2 to add a 2 to the interaction term which is why we started with HI = 2 gφAφB. If all the particles were indistinguishable (say for a self-interacting φ3 theory) we would need a factor

1 of 3! in front of our interaction term. We now plug this expression for the vev back into our original integral calculation of decay amplitude s −ig Z d3k d3k d3k 2E 2E 2E 4 1 2 A 1 2 a i(k1+k2+kA)·x M = d x 3 3 3 e 2 (2π) (2π) (2π) 2Ek1 2Ek2 2Eka

3 3 3 3 3 3 3 3 × (2π) δ (p2 − k1)(2π) δ (p1 − k2) + (2π) δ (p1 − k1)(2π) δ (p2 − k2)

3 3 × (2π) δ (kA − pA).

The integral over kA may be calculated in a manner similar to those calculated in the decay amplitude in ABC-theory, using the property of δ-functions: s Z d3k 2E A A −ikA·x 3 −ipA·x 3 e (2π) δ(pA − kA) = e . (5.50) (2π) 2EkA

Plugging this in we are left with a sum of a product of coupled integrals over k1 and k2, respectively: −ig Z M = d4xe−ipA·x 2 s s h Z d3k 2E Z d3k 2E i 1 1 ik1·x 3 3 2 2 ik2·x 3 3 × 3 e (2π) δ (p2 − k1) 3 e (2π) δ (p1 − k2) (2π) 2Ek1 (2π) 2Ek2 s s h Z d3k 2E Z d3k 2E i 1 2 ik1·x 3 3 2 2 ik2·x 3 3 + 3 e (2π) δ (p1 − k1) 3 e (2π) δ (p2 − k2) . (2π) 2Ek2 (2π) 2Ek2 (5.51) Once more we apply the rules of δ-functions to obtain Z −ig nE1 E2 o M = d4xe−ipA·x eip2·x eip1·x + eip1·xeip2·x . (5.52) 2 E2 E1

50 Combining the exponentials and cancelling the E1 and E2

−ig Z M = d4x2ei(p1+p2−pA)·x. (5.53) 2

1 Notice that we get an extra factor of 2 in the integrand which is why the 2 was included in 1 2 the interaction term of H, Hint = 2 gφAφB. Using the Fourier Transform of the exponential, we obtain as our ﬁnal result

4 4 M = −ig(2π) δ (p1 + p2 − pA). (5.54)

51 CHAPTER 6

DIMENSIONAL REGULARIZATION

The previous calculations have been for tree-level diagrams, ones with no internal

loops. When we calculate higher-order contributions to the perturbative expansion, these internal line will cause the resulting integrals to diverge. In order to make physical sense of these inﬁnite values, we introduce a process called dimensional regularization, a process

originally found in [14]. The basic idea is as follows. First we perform the calculation in D “continuous” dimensions. Note that we are in no way attempting to give a physical interpretation to the

idea of a continuous dimension. It is simply a mathematical device we use to isolate the divergence in the integral. After an application of Feynman’s Trick to the integrand, we simplify the denominator and perform a shift in the integration variable which results in a

radial integral. After integrating out the angular dependance and evaluating the remaining radial integral, we then let D = N − where N is the number of (physical) dimensions being considered and expand everything in a power series of . We will obtain a singular part, a

finite part, and a vanishing contribution. Ultimately, the singular part of the integral will be removed through a process known as renormalization. See [10], vol. II of [13], part II of [4], Ch. 9 of [16], and Ch’s. 12 and 16 of [15] for introductory treatments. By redefining L to include a “counterterm” which has a renormalization constant equal to the singular term, we can derive a new Feynman rule which will cancel out the singular part. Thus we are left with a finite result for the higher order contribution to the scattering amplitude. 6.1 One-loop Correction to the A-Propagator in ABB Theory

Consider now the “self-energy” correction to the propagator in a theory given by a cubic interaction (φ3, ABC, or ABB) as shown in Figure 6.1. Physically, this diagram is interpreted as an A-particle propagating in space, splitting into two ”virtual” particles which

52 then recombine. We perform the 2nd-order contribution in a continuous D dimensions in

order to isolate the singularity. The integral corresponding to the one-loop correction to the A-propagator in ABB theory is given by

k−q

q A q B B B

Figure 6.1: Self-Energy of The A-Particle in ABB Theory

2 Z D 2 g 1 d k 1 Σ(q ) = i D 2 2 2 2 . (6.1) 2 4 (2π) (k − mB)((k − q) − mA)

The integrand in (6.1) may be expressed as an integral via Feynmann’s trick

1 Z 1 1 = dx 2 , (6.2) AB 0 [Ax + B(1 − x)] so that (6.1) simpliﬁes after inserting this result:

2 Z D Z 1 2 g d k 1 Σ(q ) = i D dx 2 2 2 2 2 8 (2π) 0 [(k − mB)(1 − x) + ((k − q) − mA)x] g2 Z dDk Z 1 1 = i D dx 2 2 2 2 2 . 8 (2π) 0 [k − 2k · qx + q x − mB(1 − x) − mAx]

Next we complete the square of the terms containing k by adding and subtracting a q2x2 term. We then perform a change of variables k0 = k − qx. The dx and dDk0 integrals are

iterated integrals and can be ﬂipped so that we are left with

2 Z 1 Z D 0 2 g d k 1 Σ(q ) = i dx D 0 2 2 2 2 2 . (6.3) 8 0 (2π) [(k ) − q x(1 − x) − mB(1 − x) − mAx]

53 The dDk0 integral is then given by

Z dDk 1 i Γ(a − D/2) 1 a−D/2 = , (6.4) (2π)D [X2 − k2]a (4π)D/2 Γ(a) X2

2 2 2 2 so we obtain for our particular case (X = q x(1 − x) − mB(1 − x) − mAx and a = 2) 2 Z 1 2 g Γ(2 − D/2) 1 Σ(q ) = − D/2 dx 2 2 2−D/2 . (6.5) 8(4π) Γ(2) 0 [q x(1 − x) − M ] 2 2 2 M ≡ mB(1 − x) + mAx is a term that will change depending on the propagator being computed in the theory being considered. Notice that upon a change of variables y = 1 − x we recover the same integral Z 1 1 Z 1 1 dx 2 2 2 = dy 2 2 2 , (6.6) 0 q x(1 − x) − mB(1 − x) − mAx 0 q y(1 − y) − mBy − mA(1 − y) so that the choice of parameterization is arbitrary. Letting D = 4 − we will ultimately expand (6.5) in a power series of :

Z 1 2 /2 2 g g h 4πµ i Σ(q ) = 2 Γ dx 2 2 8 (4π) 2 0 q x(1 − x) − M where we made the shift g → gµ/2 to keep the integrand unitless. Now the series expansion for Γ(z) is well known for any complex z, 1 Γ(z) = − γ + O(z), (6.7) z For the integrand in (6.1) we use the identity 1 z = e log(z) = 1 + log(z) + ( log(z))2 + ··· , (6.8) 2 so that upon insertion of these series expansions, (6.1) becomes

2 Z 1 2 2 2 1−g 2 h q x(1 − x) − M 2 i Σq = − γ + O() d 1 − log 2 + O( ) . 8 4π 0 2 4πµ Our ﬁnal result for the one-loop correction to the B-propagator in ABB theory is

2 Z 1 2 2 2 1gµ h2 q x(1 − x) − M i Σ(q ) = − − γ − dx log 2 + O() (6.9) 8 4π 0 4πµ Once we introduce a new Feynman rule for self-interactions in the ABB theory (by the renormalization process discussed above), the oﬀending singular term in the expansion above will vanish and we will be left with a ﬁnite result.

54 6.2 One-loop Correction to the Vertex in ABC Theory

We have previously calculated the ﬁrst order contribution to the vertex of ABC

theory, namely the amplitude corresponding to the vertex itself. If we were to calculate the 2nd-order contribution

1 Z d4x d4x h0|cφ φ φ φ φ φ a†b†|0i, (6.10) 2 1 2 Ax Bx Cx Ay By Cy the vev vanishes since there are an odd number of ﬁelds inside it. So we continue to the third-order contribution which is shown in the following diagram: If we write the third-order

p B B q

C A A q−p B p A B q+p A C p C

Figure 6.2: One-loop correction to the vertex of ABC theory

contribution to this process we obtain (neglecting normalizations)

(−ig)3 Z d4x d4x d4x h0|ab φ φ φ φ φ φ φ φ φ c†|0i (6.11) 3! 1 2 3 Ax Bx Cx Ay By Cy Az Bz Cz

After proper manipulations which amount to applying the Feynman Rules for the theory we

are left with the remaining integral for our amplitude

Z D 3 d q 1 (−ig) D 2 2 2 2 2 2 . (6.12) (2π) [(q + pA) − mB][(q − pB) − mA](q − mC ) Using Feynman’s trick we obtain for the integrand

Z 1 1 dxdy . (6.13) h i3 0 2 2 2 2 2 2 [(q − pB) − mA]x + [(q + pA) − mB]y + (q − mC )(1 − x − y)

55 To simplify the denominator we expand everything out and complete the square then perform

0 a change of variables q = q − (pBx − pAy)

3 h 2 2 2 2 2 2 i q − 2q · (pBx − pAy) + pBx + pA − mAx − mBy − mC (1 − x − y) (6.14) 2 h 2 2 2 2i = (q − (pBx − pAy)) − pBx(1 − x) − pAy(1 − y) − M (6.15)

= [(q0)2 − P 2 − M 2]3, (6.16)

so that 6.12 becomes

Z 1 Z D 0 3 d q 1 (−ig) dxdy D 0 2 2 2 3 . (6.17) 0 (2π) [(q ) − P − M ] We can now evaluate the dDq0 integral by methods of [16] (see Ch. 8) and get

(−g)3 Z 1 Γ(3 − D/2) 1 D/2 dxdy 2 2 3−D/2 . (6.18) (4π) 0 Γ(3) [P + M ] (6.18) clearly has a form similar to the same integral done for the self-energy of the B- propagator in ABB theory. The only diﬀerence here is the Γ(3−D/2) instead of a Γ(2−D/2).

This means that this integral will be ﬁnite in 4-dimensional spacetime. But we will be left with a singular term in 6 dimensions. Letting D = 6 − we obtain

1gµ/6 3 Z 1 , − Γ dxdy (6.19) 2 4π 2 0 so that the power series expansion of the one-loop correction to the vertex becomes

1gµ/6 3h2 P 2 + M 2 i M = − − γ − log + O() . (6.20) 2 4π 4πµ2

We obtain a result similar to the one-loop correction to the propagator, but in 6 spacetime-

dimensions.

6.3 Fourth-Order Self Energy Diagram: Nested Divergence

Next we calculate the fourth-order contribution to the self-energy propagator in a cubic interaction. We will calculate this for the A-propagator in ABC theory, and we will

show that different diagrams will give different resulting contributions to the finite part which is recovered after renormalization.

56 Consider the propagator correction corresponding to an A-particle splitting into a B

and C particle. The virtual B particle then splits itself into virtual A and C particles before recombining into a B particle which then recombines with the original virtual C particle. Since there are two loops we will be integrating over two independant momentum variables.

C q B B

k A k A k−q A p p p−k C

Figure 6.3: Two-loop correction to the A-propagator: nested divergence

Let the incoming momentum of the A-particle be denoted by q. By four-momentum conservation enforced by the Feynman rules, the momentum of the outgoing virtual B-particle will

be denoted as k; for the C-particle, k + q. For the nested loop the momentum of the virtual A is denoted p and the virtual B, p + k. The resulting integral is

Z D D 4 d k d p 1 g D D 2 2 2 2 2 2 2 2 2 . (6.21) (2π) (2π) [(p − k) − mC ](k − mB) (q − mC )[(q − k) − mA]

One can clearly see that the dDp can be evaluated ﬁrst and it is the same calculation as that of the one-loop correction to the self-energy. We thus replace it with the computed values

57 expressed in terms of D (not yet in a power series of ):

Z D Z D 4 d k 1 d q 1 M = g D 2 2 2 2 2 D 2 2 2 2 (2π) [(p − k) − mC ](k − mB) (2π) (q − mC )[(q − k) − mA] Z D 4 d k 1 1 Γ(2 − D/2) = g D 2 2 2 2 2 D/2 (2π) [(p − k) − mC ](k − mB) (4π) Γ(2) Z 1 1 × dx 2 2 2 2−D/2 0 [k x(1 − x) − mC (1 − x) − mAx] Z 1 4 Γ(2 − D/2) D/2−2 = g D/2 dx[x(1 − x)] (4π) 0 Z dDk 1 × . D h 2 2 i2−D/2 (2π) 2 2 2 2 2 2 mC mA [(p − k) − mC ](k − mB) k − x − 1−x In the last line we collected the terms in the new integrand and factored out the [x(1 − x)]D/2−2. Using Feynman’s Trick, we obtain for the new integrand

Γ(5 − D/2) Z 1 y1−D/2z dydz , n 2 2 h i o5−D/2 Γ(2 − D/2) 0 2 mC mA 2 2 2 2 k − x − 1−x y + (k − mB)z + (p − k) − mC (1 − y − z) so that the original integral becomes

Z 1 Z 1 Z 1 4 Γ(5 − D/2) 2−D/2 1−D/2 M = g D/2 dx[x(1 − x)] dy y dz z (4π) 0 0 0 Z dDk 1 × . D n 2 2 h i o5−D/2 (2π) 2 mC mA 2 2 2 2 k − x − 1−x y + (k − mB)z + (p − k) − mC (1 − y − z) To evaluate the dDk integral we once again complete the square in k and then make a change of variables k0 = k − p(1 − y − z) so that the resulting integral becomes

Z dDk0 1 , (6.22) (2π)D h i5−D/2 (k0)2 + p2(1 − y − z)(y + z) − M 2 where M 2 is used to collect the mass terms and will change depending on both the theory and the particular loop. It is given by

y y M 2 = m2 + m2 − m2 z − m2 (1 − y − z). (6.23) C x A 1 − x B C

The integral (6.22) can then be evaluated accordingly and we obtain

1 Γ(5 − D) 1 . (4π)D/2 Γ(5 − D/2) (p2(1 − y − z)(y + z) − M 2)5−D

58 Inserting this result into our original integral we then obtain

Z 1 Z 1 Z 1 4 Γ(5 − D) 2−D/2 1−D/2 1 M = g D dx[x(1−x)] dy y dz z 2 2 5−D . (4π) 0 0 0 (p (1 − y − z)(y + z) − M ) (6.24) From this we can conclude that the amplitude corresponding to this process is ﬁnite in 4

dimensions of spacetime. For if we insert D = 4 − , the Γ-function would yield a ﬁnite expansion of Γ(1 + /2).

6.4 Fourth-Order Self-Energy Diagram: Overlapping Divergence

The other contribution to the self-energy at fourth-order is one where a particle splits into two virtual particles which then exchange another virtual particle before the original virtual particles recombine into the original particle as indicated by the following ﬁgure. We

p+q/2 k+q/2 q q k−p

k−q/2 p−q/2

Figure 6.4: Two-loop correction to the propagator: overlapping divergence

are going to work in φ3 theory. And in order to evaluate this type of integral, we will set m = 0. This discussion follows [15], pp. 541-546. The entire integral looks like

Z dDk dDq 1 Σ = (−ig)4 . (6.25) ov D D h i (2π) (2π) 1 2 1 2 1 2 1 2 2 (p − 2 q) (p + 2 q) (k − 2 q) (k + 2 q) (k − p)

59 We will evaluate the dDk integral ﬁrst. Expressing the denominator as an integral of Feynman

parameters we have

Z 1 Z 1−x dDk 1 − 2ig dx dy . (6.26) D h i3 0 0 (2π) 1 2 1 2 2 (k + 2 q) x + (k − 2 q) y + (k − p) (1 − x − y) The term which is cubed in the denominator of the integrand can be simpliﬁed to

1 1 k2 − 2k · q q(x − y) − p(1 − x − y) + q2(x + y) + p2(1 − x − y). (6.27) 2 4

1 So we shift the k variable k = k − 2 q(x − y) − p(1 − x − y) to complete the square, and we’re left with Z 1 Z 1−x D 2 d k 1 − 2ig dx dy D 2 2 3 , (6.28) 0 0 (2π) [k − X ] where we deﬁne X2 from

1 1 X2 = [p(1 − x − y) − q(x − y)]2 − p2(1 − x − y) − q2(x + y). (6.29) 2 4

If we make the following change of variables

1 ξ = x + y ξη = (x − y), (6.30) 2

then (6.29) becomes

1 X2 = [p(1 − ξ) − ξηq]2 − p2(1 − ξ) − q2ξ 4 1 = p2(1 − ξ)ξ − 2p · q(1 − ξ)ξη − q2(1 − ξ)ξη2 + q2η2(1 − ξ)ξ + ξ2η2q2 − q2ξ 4 1 = −(1 − ξ)ξ(p + qη)2 − q2ξ( − η2) 4 ≡ (1 − ξ)ξY 2.

From this, Y 2 can be expressed as

1 − η2 Y 2 = q2 4 − (p + qη)2. (6.31) 1 − ξ

This change of variables from x, y to ξ, η imposes the following Jacobian

∂x ∂x 1 ∂(x, y) ∂ξ ∂η 2 + η ξ = = = ξ, ∂(ξ, η) ∂y ∂y 1 ∂η ∂η 2 − η −ξ

60 which also changes the integration of the Feynman parameters

Z 1 Z 1 Z 1 Z 1/2 dx dy ⇒ ξdξ dη. 0 0 0 −1/2

Our dDk integral then becomes

Z 1 Z 1/2 Z D 2 Z 1 Z 1/2 2 3−D/2 2 d k 1 2g Γ(3 − D/2) µ −2ig ξdξ dη D 2 2 3 = D/2 ξdξ dη 2 . 0 −1/2 (2π) [k − X ] (4π) Γ(3) 0 −1/2 X

Setting D = 6 − we obtain our result for the dDk integral in terms of . Below is written

D the singular part Λ0(p, q) and the ﬁnite contribution ΛR(p, q) of the d k integral:

g2 Z 1 Z 1/2 X2 ΛR(p, q) = − 3 ξdξ dη log 2 (6.32) 2(4π) 0 −1/2 µ g2 1 Λ = Γ 1 + . (6.33) 0 (4π)D/2 2

We use the integral for Λ(p, q) for the full integral Σov. Recalling that the momentums

1 for the other two virtual particles are (p ± 2 q, our full integral reads

g2 Z dDp Λ(p, q) Σ = i (6.34) ov D 1 2 1 2 2 (2π) (p + 2 q) (p + 2 q) ig4 Z 1 Z 1/2 Z dDp ξ1−/2(1 − ξ)−/2(µ2) = Γ dξ dη , (6.35) D/2 D 1 2 1 2 2 −/2 2(4π) 2 0 −1/2 (2π) (p + 2 q) (p + 2 q) (Y )

where in the second line we combined the ξ from the Jacobian with the [(1 − ξ)ξ)]−/2 from the X2 term leaving Y 2.

Because of the overlapping divergence we must resort to improving the convergence of the dDp integral by following a process similar to the original one in [14]. It amounts to an integration by parts of

Z 1/2 1 dη . (6.36) 1 2 1 2 2 −/2 −1/2 (p + 2 q) (p − 2 q) (Y )

We ﬁrst set

1 1 u = dv = (6.37) 2 /2 1 2 1 2 (Y ) (p − 2 q) (p + 2 q)

61 which gives the terms in the new integrand:

1 ∂Y 2 1 du = v = . 2 1+/2 1 2 1 2 2 (Y ) ∂η (p − 2 q) (p + 2 q)

∂Y 2 The calculation of ∂η is straightforward:

∂Y 2 2ηq2 = −2p · q − 2ηq2 + . ∂η 1 − ξ

These results give us our integration by parts:

Z 1/2 1 dη 1 2 1 2 2 /2 −1/2 (p − 2 q) (p + 2 q) (Y ) 2 2ηq2 1 1 1/2 Z 1/2 −2p · q − 2ηq + 1−ξ = 1 1 − ηdη 1 1 . (6.38) 2 /2 2 2 −1/2 2 2 2 1+/2 (Y ) (p + 2 q) (p + 2 q) 2 −1/2 (p − 2 q) (p + 2 q) (Y ) 6.4.1 Calculating the Boundary Contribution of the dη Integral

1 2 For the boundary term, setting η = ± 2 into (6.31) cancels the q term leaving (p ± 1 2 2 q) , and we are left with

1 1 1/2 1h 1 1 i

1 1 = 1 1 + 1 1 . 2 /2 2 2 −1/2 1+/2 2 2 1+/2 (Y ) (p + 2 q) (p + 2 q) 2 (p + 2 q) (p − 2 q) (p + 2 q) (p − 2 q) (6.39)

The ﬁrst part of Σov from the boundary term can then be calculated:

ig4 Z 1 Z dDp (µ2) Σ(1) = Γ ξ1−/2(1 − ξ)−/2 . (6.40) ov D/2 D 1 2 1+/2 1 2 2(4π) 2 0 (2π) [(p + 2 q) ] (p − 2 q) For the dξ integral note that ξ1−/2(1 − ξ)−/2 = ξ(2−/2)−1(1 − ξ)(1−/2)−1 which is exactly the form we need for the deﬁnition of the B-function

Z 1 Γ(ρ)Γ(σ) B(ρ, σ) ≡ dz zρ−1(1 − z)σ−1 = . (6.41) 0 Γ(ρ + σ)

To calculate the dDp integral we use Feynman parameters as usual. The denominator in the integrand is expressed as

1 Γ((2 + /2) Z 1 x/2 = dx (6.42) 1 1+/2 1 2 h i2+/2 (p + 2 q) (p − 2 q) Γ(1 + /2) 0 1 2 1 2 (p + 2 q) x + (p − 2 q) (1 − x)

62 which is the same expression for the other boundary term. The term in the denominator raised to the 2 + /2 power can be simpliﬁed to

1 1 1 (p + q)2x + (p − q)2(1 − x) = p2 − p · q(1 − 2x) + q2 2 2 4 1 1 1 1 = p2 − 2p · q (1 − 2x) + q2 (1 − 2x)2 − q2 (1 − 2x)2 + q2 2 4 4 4 1 2 1h i 1 = p − q (1 − 2x) − q2 − 4q2x + 4q2x2 + q2 2 4 4 = p2 + q2x(1 − x),

1 where in the last step we shifted p → 2 q(1 − 2x). Combining (6.42) and the ξ’s in (6.41) we obtain

4 Z 1 Z D 2 /2 (1) ig Γ(2 + /2) d p (µ ) x Σov = D/2 Γ B 2 − , 1 − dx D/2 2 2 2+/2 . 2(4π) 2 2 2 Γ(1 + /2) 0 (2π) [p + q x(1 − x)] (6.43) The dDp integral can ﬁnally be calculated and we obtain

Z dDp (µ2) 1 Γ(2 + /2 − D/2) (µ2) = . (6.44) (2π)D/2 [p2 + q2x(1 − x)]2+/2 (4π)D/2 Γ(2 + /2) [q2x(1 − x)]2+/2−D/2

So we plug this result into (6.43). The (4π)D/2’s combine. Noting that 2 + /2 − D/2 = − 1 when D = 6 − , we obtain a factor of q2 in the numerator. The [x(x − 1)]−1 term in the

/2 denominator combines with the x to give another B-function. Finally the Γ( 2 ) cancels 2 with the Γ(1 + 2 ) = 2 Γ( 2 ) in the denominator leaving a . Collecting all these results we are left with

g4 2 µ2 Σ(1) = q2 B 2 − , 1 − B 2 − , 2 − Γ( − 1) . (6.45) ov 2(4π)D 2 2 2 q2

2 Clearly the and Γ(−1) terms will be the ones that contribute to the singularity at the end (1) of an expansion of Σov in a series of . From [16] we know that all poles of the Γ-function (at the negative integers) are simple (see Appendix 8D of [16]):

(−1)n 1 Γ(−n + ) = + O(0) . (6.46) n!

63 By using (6.46) and ignoring the terms in the B-functions we obtain an approximation for

(6.45):

g4 2 −1µ2 Σ ≈ q2 B(2, 1)B(2, 2) (6.47) ov 2(4π)6 q2 g4 h 1 µ2 i = −q2 (6.48) 12(4π)6 2 q2 g4 h 1 1 q2 i ≈ −q2 − log + ··· . (6.49) 12(4π)6 2 µ2

6.5 Calculating the integral contribution to the dη Integral

We now turn our attention to calculating the 2nd term in (6.25) which is from the integral in (6.38). It takes the form

4 Z 1 Z D (2) ig d p 1−/2 −/2− Σov = D/2 Γ dξ D ξ (1 − ) 2(4π) 2 0 (2π) 2 2 Z 1/2 2p · q + 2ηq2 − 2ηq ηdη 1−ξ 1 2 1 2 2 1+/2 −1/2 (p + 2 q) (p − 2 q) (Y ) 4 Z 1 Z 1/2 ig 1−/2 −/2 = − D/2 Γ 1 + dξ ηdηξ (1 − ) 2(4π) 2 0 −1/2 2 2 2ηq2 D Z d p (µ ) 2p · q + 2ηq − 1−ξ . D 1 2 1 2 2 1+/2 (2π) (p + 2 q) (p − 2 q) (Y )

D 2 2 2 1/4−η2 If we make a shift in p by p → p − ηq then the d p integral becomes (Y → p + q 1−ξ )

2 2ηq2 D Z d p (µ ) 2p · q − 1−ξ I = . (6.50) D h i2h i2h 2 i1+/2 (2π) 1 1 2 2 1/4−η p + ( 2 − η)q p − ( 2 + η)q p + q 1−ξ

We apply Feynman’s trick as usual. The denominator in the integrand of (6.50) can be written as

h i2h i2h 2 i1+/2 1 1 2 2 1/4−η p + ( 2 − η)q p − ( 2 + η)q p + q 1−ξ Γ(3 + /2) Z 1 Z 1−x = dx dy(1 − x − y)/2 Γ(1 + /2) 0 0 1 . h 2 2 2 i3+/2 1 1 2 2 1/4−η p + ( 2 − η)q x + p − ( 2 + η)q y + p + q 1−ξ (1 − x − y)

64 Some algebra is required to reduce the denominator to a more useful form, and we perfom it

in detail here. The same process applies to previous simpliﬁcations: we complete the square with the term linear in p and simplify the remaining terms. 2 2 1/4 − η2 p + ( 1 − η)q x + p − ( 1 + η)q y + p2 + q2 (1 − x − y) 2 2 1 − ξ 2 h 1 1 i = p + 2p · q ( 2 − η)x − ( 2 + η)y h i 1/4 − η2 + q2 ( 1 − η)2x − ( 1 + η)2y + q2 (1 − x − y) 2 2 1 − ξ 2 2 h 1 1 i 2h 1 1 i = p + q ( 2 − η)x − ( 2 + η)y − q ( 2 − η)x − ( 2 + η)y h i 1/4 − η2 + q2 ( 1 − η)2x − ( 1 + η)2y + q2 (1 − x − y) 2 2 1 − ξ = p2 + Zq2.

1 1 In the last line we shifted p → p + q[( 2 − η)x − ( 2 + η)y] and defined Z by 1 1 1 1/4 − η2 Z = − η x(1 − x) + + η y(1 − y) + − η2 2xy + (1 − x − y) 2 2 4 1 − ξ 1 = x(1 − x) + y(1 − y) + 2xy − η x(1 − x) − y(1 − y) 4 1/4 − η2 + η2 x(1 − x) + y(1 − y) − 2xy + (1 − x − y)( 1 − η)x − ( 1 + η)y. 1 − ξ 2 2 1 Each of the first three terms can be simplified to recover the eq. (16.57) in [15]. For the 4 term we have

x(1 − x) + y(1 − y) + 2xy = x + y − (x2 + y2 − 2xy)

= x + y − (x − y)2.

For the η term we have

x(1 − x) − y(1 − y) = x(1 − x) − xy − y(1 − y) + xy

= (x − y)(1 − x − y),

and for the η2 term we have

x(1 − x) + y(1 − y) − 2xy = x(1 − x) − xy + y(1 − y) − xy

= (x + y)(1 − x − y).

65 Both the η and η2 terms contain a (1 − x − y) which can be factored out along with the one

in the last term ﬁnally obtain eq. (16.57) of [15] (see p. 544): 1 h 1/4 − η2 i Z = (x + y − (x − y)2) + η2(x + y) − η(x − y) + (1 − x − y) 4 1 − ξ b = a + Z . Z 1 − ξ

where aZ and bZ are deﬁned by 1 a = (x + y − (x − y)2) + [η2(x + y) − η(x − y)](1 − x − y) (6.51) Z 4 1 2 bZ = ( 4 − η )(1 − x − y). (6.52)

Now that we have the denominator of (6.50) in a more simpliﬁed form we can actually

compute it. One small matter, however is that the numerator was also shifted: 2ηq2 h i 2p · q − → (2q2) ( 1 − η)x − ( 1 + η)y . 1 − ξ 2 2 Note that the p · q term vanishes since it is now odd in the dDp integral (by the p2 now in the denominator.) The dDp integral can now be computed in the normal way:

Z dDp 1 1 Γ(3 + /2 − D/2) µ = i . (6.53) (2π)D [p2 + Zq2]3+/2 (4π)D/2 Γ(3 + /2) Zq2 We obtain at once the result for I:

2 Z 1 Z 1−x 2 −iΓ()(2q ) /2h1 η i µ I = D/2 dx dy (1 − x − y) (x − y) − η(x + y) + 2 . (4π) Γ(1 + /2) 0 0 2 1 + ξ Zq (6.54)

Of all the Feynman parameters that have been generated through the course of this calculation we integrate dξ by parts. Note that

µ2 µ2 µ2 = Z− = [(1 − ξ)a + b ]−(1 − ξ). Zq2 q2 q2 Z Z We can combine the (1 − ξ) term with the original one in the dξ integral. Neglecting the

ﬁrst two terms in the square bracket of (6.54), we can also absorb the other 1 − ξ terms so that the dξ integral takes the form Z 1 1−/2 − /2−1 dξ ξ [(1 − ξ)aZ + bZ ] (1 − ξ) . (6.55) 0

66 To integrate by parts we take

1−/2 − u = ξ [(1 − ξ)aZ + bZ ]

dv = (1 − ξ)/2−1dξ which give v and du by

h −/2 − 1−/2 −(1+) i du = (1 − 2 ξ [(1 − ξ)aZ + bz] + ξ [(1 − ξ)aZ + bZ ] aZ dξ 2 v = − (1 − ξ)/2.

The boundary term vanishes and so will the 2nd term in du. We are left with

Z 1 2 /2 −/2 − − dξ(1 − ξ) (1 − 2 ξ [(1 − ξ)aZ + bz] . (6.56) 0

1 Taking → 0 for everything we are left with (keeping only the 2 ) terms

g4 h 1 1 q2 i Σ(2) = −q2 − log + ··· . (6.57) ov 12(4π)6 2 µ2

67 CHAPTER 7

CONCLUSION AND FUTURE WORK

In conclusion, we set out to give as many possible details in deriving some standard

and some uncommon results in scalar quantum ﬁeld theory. For a real scalar ﬁeld φ we started with a Lagrangian (density.) We found the conjugate momentum density π then used it along with φ to determine our classical physical quantities of interest.

Next we promoted our ﬁelds to operators and imposed commutation relations between them. We expressed them as a Fourier integral of a linear combination of creation and annihilation operators. These creation and annihilation operators were deﬁned to act on multi-particle states in the Fock space, creating or destroying particles at will. When

† the Hamiltonian and momentum operators were expressed in terms of ap and ap we then showed that the energy and momentum of these states were precisely equal to the sum of

Energy/momentum of each of the particles in the state. We then introduced an interaction term in the Lagrangian. After expressing the new ground state |Omegai in terms of the original ground state |0i of the free theory we calculated probability amplitudes (or scattering amplitude) for states to evolve to different states through the interaction. These scattering amplitudes are interpreted as Feynman diagrams. Unfortunately diagrams containing internal loops led to infinities, so we regularized the diagram to isolate and remove the singularity, leaving a finite contribution. There are several ways this work could be extended. The first would be to continue with even higher-order calculations. One example would be the box diagram and its corresponding one and two-loop corrections. The calculations are straightforward, and no new techniques would be needed. It would also better highlight the aspects of renormalization. Another interesting possibility, one that I have not yet seen, would be to study interactions of distinguishable particles with quartic interactions. φ4 theory is a well-known,

68 commonly studied theory with similar calculations, yet there exist related theories such as

ABCD ABCC AABB ABBB. (7.1)

Deriving the Feynman rules, computing the symmetry factors of the diagrams, and determining renormalization constants are all potential options to pursue in this direction. Yet another direction to extend this work, and perhaps a more useful one, would be to extend our discussion to complex scalar ﬁelds. Determined from the Lagrangian (cf. eq.

(2.22))

∗ µ 2 ∗ L = ∂µφ ∂ φ − m φ φ. (7.2)

The dynamical variables φ and φ∗ (with corresponding conjugate momenta π and π∗) can be shown to satisfy the Hamiltonian (see [4], Problem 2.2, p. 33)

H = π∗π + (∇φ∗) · (∇φ) + m2φ∗φ (7.3) which is similar to H for a real scalar field. The new quantity one can compute for a complex scalar field is the charge defined by

Z i Q = d3x φ∗π∗ − φπ . (7.4) 2

∗ Once interactions are introduced (after appropriate quantization) such as Hint = gφAφBφB, then we can begin interpreting creation and annihilation of antiparticles, particles corresponding to a certain particle with the same mass but opposite charge.

As a means to gain more physical insight into these theories, we could calculate the cross-sections and decay rates corresponding to the particular amplitudes. All loop diagrams, after appropriate renormalization, ultimately yield a ﬁnite contribution to the scattering amplitude. It is my understanding that in actually physical theories these higher-order terms give a correction to the measurable quantity (the mass, for instance.) A ﬁnal direction would be to discuss the renormalization of these theories. Depending on the dimension in which the divergent integrals are regularized, a theory may be super- renormalizable, renormalizable, or non-renormalizable ([7].) In D = 4 dimensions, ABC

69 is a super-renormalizable theory, meaning it only takes a ﬁnte number of counterterms to make terms of all orders ﬁnite. The quantum theory of gravity is an example of a non- renormalizable theory.

70 REFERENCES

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